The post Factors Formula | How To Find Sum of Factors of Composite Numbers appeared first on .

]]>Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In the post of * Factors and Multiples of 1 to 100*, you get that the sum of factors of prime numbers and composite numbers are given in the table. In this article, you will learn about

Since the factors of a prime number is one and itself, the sum of factors of a prime number is one more than the prime number.

For example: let a prime number is 59. Then, its factors are 1 and 59 only. Thus, the sum of the factors of 59 is 59 + 1 = 60.

It is very simple to find sum of factors of a prime number.

Now, take another example of a composite number 150.

All factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Sum of all factors of 150 is (1 + 2 + 3 + 5 + 6 + 10 + 15 + 25 + 30 + 50 + 75 + 150) = 372.

**Read More About Factors Formulas**

How To Find Total Number Of Factor of Any Number

How To Find Number of Even Factor of Any Number

How To Find Number of Odd Factor of Any Number

Since these numbers are smaller numbers, you get the sum of factors easily by adding all the factors of that number.

But if a bigger composite number, let 4200 is given, then it is very difficult to find the sum of all factors as it is very time consuming process to list all the factors of a bigger composite numbers.

Therefore, you need a short trick to find the sum of all factors of a composite number.

Let N is the composite number whose the sum of all its factors is to be calculated.

Then, its prime factorization is N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α, β, γ, δ, … are prime numbers and a, b, c, d, … positive integers.

The factors of α^{a} are 1, α^{1}, α^{2}, α^{3}, … α^{a}. Sum of factors of α^{a} is (1 + α^{1 }+ α^{2 }+ α^{3 }+ … + α^{a}).

The factors of β^{b} are 1, β^{1}, β^{2}, β^{3}, … β^{b}. Sum of factors of β^{b} are (1 + β^{1 }+ β^{2 }+ β^{3 }+ … + β^{b}).

The factors of γ^{c} are 1, γ^{1}, γ^{2}, γ^{3}, … γ^{b}. Sum of factors of γ^{c} are (1 + γ^{1 }+ γ^{2 }+ γ^{3 }+ … + γ^{c}).

and so on…

Thus, sum of all factors of N is

S = (1 + α^{1 }+ α^{2 }+ α^{3 }+ … + α^{a}) × (1 + β^{1 }+ β^{2 }+ β^{3 }+ … + β^{b}) × (1 + γ^{1 }+ γ^{2 }+ γ^{3 }+ … + γ^{c}) × …

Using the sum formula of geometric progression, you get short trick to find sum of all factors of a composite numbers as below.

Sum of All Factors of N = α^{a} × β^{b} × γ^{c} × … is

[(α^{a+1} – 1)/(α – 1)] × [(β^{b+1} – 1)/(β – 1)] × [(γ^{c+1} – 1)/(γ – 1)] ×…

[(α

This sum of all factors of N includes 1 and the number N itself.

**Solution** : Prime Factorization of 36 is 36 = 2^{2} × 3^{2}.

Thus, the sum of all factors of 36 is [(2^{2+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)] = 91.

**Solution** : Prime Factorization of 216 is 216 = 2^{3} × 3^{3}.

Thus, the sum of all factors of 216 is [(2^{3+1} – 1)/(2 – 1)][(3^{3+1} – 1)/(3 – 1)] = 600.

**Solution** : Prime Factorization of 196 is 196 = 2^{2} × 7^{2}.

Thus, the sum of all factors of 196 is [(2^{2+1} – 1)/(2 – 1)][(7^{2+1} – 1)/(7 – 1)] = 399.

**Solution** : Prime Factorization of 60 is 60 = 2^{2} × 3 × 5.

Thus, the sum of all factors of 60 is [(2^{2+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 168.

**Solution** : Prime Factorization of 96 is 96 = 2^{5} × 3.

Thus, the sum of all factors of 96 is [(2^{5+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)] = 252.

**Solution** : Prime Factorization of 2800 is 2800 = 2^{4} × 5^{2} × 7.

Thus, the sum of all factors of 2800 is [(2^{4+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{1+1} – 1)/(7 – 1)] = 7688.

**Solution** : Prime Factorization of 16 is 16 = 2^{4}.

Thus, the sum of all factors of 16 is [(2^{4+1} – 1)/(2 – 1)] = 31.

**Solution** : Prime Factorization of 324 is 324 = 2^{2}× 3^{4}

Thus, the sum of all factors of 324 is [(2^{2+1} – 1)/(2 – 1)][(3^{4+1} – 1)/(3 – 1)] = 847.

**Solution** : Prime Factorization of 240 is 240 = 2^{4 }× 3^{ }× 5

Thus, the sum of all factors of 240 is [(2^{4+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 744.

**Solution** : Prime Factorization of 91476 is 91476 = 2^{2} x 3^{3} x 7^{1} x 11^{2}

Thus, the sum of all factors of 91476 is [(2^{2+1} – 1)/(2 – 1)][(3^{3+1} – 1)/(3 – 1)][(7^{1+1} – 1)/(7 – 1)][(11^{1+1} – 1)/(11 – 1)] = 297,920.

**Solution** : Prime Factorization of 496 is 496 = 2^{4 }× 31

Thus, the sum of all factors of 496 is [(2^{4+1} – 1)/(2 – 1)][(31^{1+1} – 1)/(31 – 1)] = 992.

**Solution** : Prime Factorization of 56 is 56 = 2^{3 }× 7

Thus, the sum of all factors of 56 is [(2^{3+1} – 1)/(2 – 1)][(7^{1+1} – 1)/(7 – 1)] = 120.

**Solution** : Prime Factorization of 19600 is 19600 = 2^{4} x 5^{2} x 7^{2}

Thus, the sum of all factors of 19600 is [(2^{4+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{2+1} – 1)/(7 – 1)] = 54777.

**Solution** : Prime Factorization of 100 is 100 = 2^{2} x 5^{2}

Thus, the sum of all factors of 100 is [(2^{2+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)] = 217.

**Solution** : Prime Factorization of 48 is 48 = 2^{4} x 3

Thus, the sum of all factors of 48 is [(2^{4+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)] = 124.

**Solution** : Prime Factorization of 360 is 360 = 2^{3} x 3^{2} x 5^{1}.

Thus, the sum of all factors of 360 is [(2^{3+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 1170.

**Solution** : Prime Factorization of 25 is 25 = 5^{2}.

Thus, the sum of all factors of 25 is [(5^{2+1} – 1)/(5 – 1)] = 31.

**Solution** : Prime Factorization of 2450 is 2450 = 2^{1} x 5^{2} x 7^{2}.

Thus, the sum of all factors of 2450 is [(2^{1+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{2+1} – 1)/(7 – 1)] = 5301.

**Solution** : Prime Factorization of 720 is 720 = 2^{4} x 3^{2} x 5^{1}.

Thus, the sum of all factors of 720 is [(2^{4+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 2418.

**Solution** : Prime Factorization of 243 is 243 = 3^{5}.

Thus, the sum of all factors of 243 is [(3^{5+1} – 1)/(3 – 1)] = 364.

**Solution** : Prime Factorization of 120 is 120 = 2^{3} x 3^{1} x 5^{1}.

Thus, the sum of all factors of 120 is [(2^{3+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 360.

**Solution** : Prime Factorization of 576 is 576 = 2^{6} x 3^{2}.

Thus, the sum of all factors of 576 is [(2^{6+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)] = 1651.

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]]>The post Factor Formula | How To Find Number of ODD Factors of Any Number appeared first on .

]]>To understand what is going to be find, lets start with an example.

Suppose you are given a number 1250.

If N is the number whose prime factorization is

N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α, β, γ, δ, … are prime numbers and a, b, c, d, … positive integers.

Then the total number of factors of N is (a + 1)(b + 1)(c + 1)(d + 1)….

Here, you are given N = 1250 (an even number).

Then, its **prime factorization is 1250 = ****2 × 5 ^{4}.** …………………………………………(1)

Thus, Total number of factors of 1250 is (1 + 1)(4 + 1) = 10

Now, the factors of 1250 are 1, 2, 5, 10, 25, 50, 125, 250, 625, 1250.

How these factors are calculated, look below the pattern:

Since the prime factorization of 1250 is **2 × 5 ^{4 }= 2 × 5 × 5 × 5 × 5.**

1 × 1 = 1 (odd)

1 × 2 = 2 (even)

1 × 5 = 5 (odd)

2 × 5 = 10 (even)

5 × 5 = 25 (odd)

2 × 5 × 5 = 50 (even)

5 × 5 × 5 = 125 (odd)

2 × 5 × 5 × 5 = 250 (even)

5 × 5 × 5 × 5 = 625 (odd)

2 × 5 × 5 × 5 × 5 = 1250 (even)

In these factors, you found that 1, 5, 25, 125, 625 are the ODD factors of 1250. There are only 5 odd factors of 1250.

These ODD factors are obtained if the even prime factor 2 is excluded in the product, otherwise you will get an even factor.

Let another N = 135 (an odd number).

Then, its **prime factorization is 135 = ****3 ^{3} × 5.** ……………………………………..(2)

Thus, Total number of factors of 135 is (3 + 1)(1 + 1) = 4 × 2 = 8.

Now, the factors of 135 are 1, 3, 5, 9,15, 27, 45, 135.

How these factors are calculated, look below the pattern:

Since the prime factorization of 135 is** ****3 ^{3} × 5 = **

1 × 1 = 1 (odd)

1 × 3 = 3 (odd)

1 × 5 = 5 (odd)

3 × 3 = 9 (odd)

3 × 5 = 15 (odd)

3 × 3 × 3 = 27 (odd)

3 × 3 × 5 = 45 (odd)

3 × 3 × 3 × 5 = 135 (odd)

In these factors, you found that 1, 3, 5, 9, 15, 27, 45, 135, all are the ODD factors of 135. There are only 8 odd factors of 135.

In the prime factorization of 135, there is NO even prime factor 2. So you get all factors are ODD.

Since 1250 and 135 is not a large number, so you can find odd factors of 1250 and 135 easily by listing all the factors.

BUT, if you are given a large number whose the number is factors is large, then you find that it is very difficult to find number of odd factors of the number.

Therefore, here is given factor formula, using that you can find the number of odd factors of the number, without listing all the factors.

From these two above example, you can conclude that if an odd number has all ODD factors whereas an even number can have odd and even factors. How, you can say?

Observe the prime factorizations of 1250 and 135.

Since the prime factorization of 1250 contains the prime factor 2, so 1250 has even factor and odd factor both. To find an odd factor, you need to exclude the even prime factor 2.

whereas, the prime factorization of 135 does not contain the prime factor 2, so 135 has no even factors, all factors are odd.

Thus, the number of odd factors depends on the prime factor 2 of prime factorization of any number.

It means only an even number can have an even and an odd factors whereas an odd number has only odd factors.

Now, understand the factor formula below to find the number of odd factors of any number.

Let N is the number whose the prime factorization is N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α < β < γ < δ < … are prime numbers and a, b, c, d, … positive integers.

Total Number of ODD Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is

**(b + 1)(c + 1)(d + 1)…**, if α = 2 (N is an EVEN number)

Total Number of ODD Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is

**(a + 1)(b + 1)(c + 1)(d + 1)…**, if α **≠** 2 (N is an ODD number)

**Ques 1** : Find the total number of odd factors of 120.

**Solution** : Prime Factorization of 120 is 120 = 2^{3} × 3^{1} × 5^{1}.

Thus, Total number of odd factors of 120 is (1 + 1)(1 + 1) = 2 × 2 = 4.

**Ques 2** : Find the total number of odd factors of 84.

**Solution** : Prime Factorization of 84 is 84 = 2^{2} × 3^{1} × 7^{1}.

Thus, Total number of odd factors of 84 is (1 + 1)(1 + 1) = 2 × 2 = 4.

**Ques 3** : What is the number of odd factors of the number 3600?

**Solution** : Prime Factorization of 3600 is 3600 = 2^{4} × 3^{2} × 5^{2}.

Thus, Total number of odd factors of 3600 is (2 + 1)(2 + 1) = 3 × 3 = 9.

**Ques 4** : What is the number of odd factors of the number 504?

**Solution** : Prime Factorization of 504 is 504 = 2^{3} × 3^{2} × 7^{1}.

Thus, Total number of odd factors of 504 is (2 + 1)(1 + 1) = 3 × 2 = 6.

**Ques 5** : Find the total number of odd factors of 180.

**Solution** : Prime Factorization of 180 is 180 = 2^{2} × 3^{2} × 5^{1}.

Thus, Total number of odd factors of 180 is (2 + 1)(1 + 1) = 3 × 2 = 6.

**Ques 6** : What is the number of odd factors of 6480?

**Solution** : Prime Factorization of 6480 is 6480 = 2^{4} x 3^{4} x 5^{1}.

Thus, Total number of odd factors of 6480 is (4 + 1)(1 + 1) = 5 × 2 = 10.

**Ques 7** : What is the number of odd factors of 420?

**Solution** : Prime Factorization of 420 is 420 = 2^{2} x 3^{1} x 5^{1 }x 7^{1}.

Thus, Total number of odd factors of 420 is (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8.

**Read** : Get Factors and Multiples of 1 to 100

How To Find Sum of Factors of Any Number

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]]>The post Factor Formula | How To Find Number of EVEN Factors of Any Number appeared first on .

]]>To understand what is going to be find, lets start with an example.

Suppose you are given a number 150.

If N is the number whose prime factorization is

N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α, β, γ, δ, … are prime numbers and a, b, c, d, … positive integers.

Then the total number of factors of N is (a + 1)(b + 1)(c + 1)(d + 1)….

Here, you are given N = 150 (an even number).

Then, its **prime factorization is 150 = ****2 × 3 × 5 ^{2}.** …………………………………………(1)

Thus, Total number of factors of 150 is (1 + 1)(1 + 1)(2 + 1) = 2 × 2 × 3 = 12.

Now, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

In these factors, you found that 2, 6, 10, 30, 50, 150 are the EVEN factors of 150. There are only 6 even factors of 150.

Let another N = 6655 (an odd number).

Then, its **prime factorization is 6655 = ****5 × 11 ^{3}.** ……………………………………..(2)

Thus, Total number of factors of 6655 is (1 + 1)(3 + 1) = 2 × 3 = 6.

Now, the factors of 6655 are 1, 5, 11, 55, 121, 605, 1331, 6655.

In these factors, you found that there is no EVEN factors of 6655.

Since 150 and 6655 is not a large number, so you can find even factors of 150 and 6655 easily by listing all the factors.

BUT, if you are given a large number whose the number is factors is large, then you find that it is very difficult to find number of even factors of the number.

Therefore, here is given factor formula, using that you can find the number of even factors of the number, without listing all the factors.

Have you observed that 150 has 6 even factors whereas 6655 has no even factors? Why does this happen?

Observe the prime factorizations of 150 and 6655.

Since the prime factorization of 150 contains the prime factor 2, so 150 has even factor.

whereas, the prime factorization of 6655 does not contain the prime factor 2, so 6655 has no even factors.

Thus, the number of even factors depends on the prime factor 2 of prime factorization of any number.

It means only an even number has even factors whereas an odd number has no even factors.

Now, understand the factor formula below to find the number of even factors of any number.

Let N is the number whose the prime factorization is N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α < β < γ < δ < … are prime numbers and a, b, c, d, … positive integers.

Total Number of EVEN Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is

**(a)(b + 1)(c + 1)(d + 1)…**, if α = 2 (N is an EVEN number)

Total Number of EVEN Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is

**0**, if α **≠** 2 (N is an ODD number)

**Ques 1** : Find the total number of even factors of 120.

**Solution** : Prime Factorization of 120 is 120 = 2^{3} × 3^{1} × 5^{1}.

Thus, Total number of even factors of 120 is (3)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12.

**Ques 2** : Find the total number of even factors of 84.

**Solution** : Prime Factorization of 84 is 84 = 2^{2} × 3^{1} × 7^{1}.

Thus, Total number of even factors of 84 is (2)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8.

**Ques 3** : What is the number of even factors of the number 3600?

**Solution** : Prime Factorization of 3600 is 3600 = 2^{4} × 3^{2} × 5^{2}.

Thus, Total number of even factors of 3600 is (4)(2 + 1)(2 + 1) = 4 × 3 × 3 = 36.

**Ques 4** : What is the number of even factors of the number 504?

**Solution** : Prime Factorization of 504 is 504 = 2^{3} × 3^{2} × 7^{1}.

Thus, Total number of even factors of 504 is (3)(2 + 1)(1 + 1) = 3 × 3 × 2 = 18.

**Ques 5** : Find the total number of even factors of 180.

**Solution** : Prime Factorization of 180 is 180 = 2^{2} × 3^{2} × 5^{1}.

Thus, Total number of even factors of 180 is (2)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12.

**Ques 6** : What is the number of even factors of 6480?

**Solution** : Prime Factorization of 6480 is 6480 = 2^{4} x 3^{4} x 5^{1}.

Thus, Total number of even factors of 6480 is (4)(4 + 1)(1 + 1) = 4 × 5 × 2 = 40.

**Ques 7** : What is the number of even factors of 420?

**Solution** : Prime Factorization of 420 is 420 = 2^{2} x 3^{1} x 5^{1 }x 7^{1}.

Thus, Total number of even factors of 420 is (2)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16.

**Read** : Get Factors and Multiples of 1 to 100

**Read** : How to Find Number of ODD Factors of Any Number

**Read** : How To Find Sum of Factors of Any Number

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]]>According to fundamental theorem of arithmetic, every positive integer can be expressed as the product of primes in a unique way. OR, you can also say that a number can be resolved into prime factors in only one way.

It mean, for a positive integer N > 1 can be uniquely written as,

N = α^{a} × β^{b} × γ^{c} × δ^{d }× …

where α < β < γ < δ < … are prime numbers and a, b, c, d, … are integer greater than or equal to 0. This decomposition of N is also known as prime factorization of N.

Let N denotes a number whose prime factorization

N = p_{1}p_{2}p_{3}…, where p_{1, }p_{2, }p_{3, }… are prime numbers.

Suppose N is also decomposed as N = q_{1}q_{2}q_{3}…, where q_{1, }q_{2, }q_{3, }… are other prime numbers.

Then, p_{1}p_{2}p_{3}… = q_{1}q_{2}q_{3}…,

Now, p1 divides the product of p_{1}p_{2}p_{3}… and since each of the factors of this product is a prime, therefore q_{1 }divides one of them p_{1}p_{2}p_{3}…,

Let q_{1 }divides one of them p_{1}. But p_{1 }and q_{1 }are prime, therefore p_{1 }and q_{1 }are equal.

Thus, p_{1}p_{2}p_{3}… = q_{1}q_{2}q_{3}…, ⇒ p_{2}p_{3}… = q_{2}q_{3}…,

Similarly q_{2 }divides one of them p_{2}p_{3}…,

Let q_{2 }divides one of them p_{2}. But p_{2 }and q_{2 }are prime, therefore p_{2 }and q_{2 }are equal.

Therefore, the prime factorization p_{1}p_{2}p_{3}… are q_{1}q_{2}q_{3}… same.

Hence, N can be resolved into prime factors in one way.

Let N is the number whose number of factors is to be calculated and its prime factorization is N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α, β, γ, δ, … are prime numbers and a, b, c, d, … positive integers.

Now, the factors of α^{a} are 1, α^{1}, α^{2}, α^{3}, … α^{a}. Total number of factors of α^{a} is (a + 1).

The factors of β^{b} are 1, β^{1}, β^{2}, β^{3}, … β^{b}. Total number of factors of β^{b} is (b + 1).

Similarly,

Total number of factors of γ^{c} is (c + 1).

Total number of factors of δ^{d} is (d + 1).

Thus,

Total Number of Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is

(a + 1)(b + 1)(c + 1)(d + 1)…

(a + 1)(b + 1)(c + 1)(d + 1)…

This total number of factors of N includes 1 and the number N itself.

**Ques 1** : Find the total number of factors of 120.

**Solution** : Prime Factorization of 120 is 120 = 2^{3} × 3^{1} × 5^{1}.

Thus, Total number of factors of 120 is (3 + 1)(1 + 1)(1 + 1) = 4 × 2 × 2 = 16.

**Ques 2** : Find the total number of factors of 84.

**Solution** : Prime Factorization of 84 is 84 = 2^{2} × 3^{1} × 7^{1}.

Thus, Total number of factors of 84 is (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12.

**Ques 3** : What is the number of factors of the number 3600?

**Solution** : Prime Factorization of 3600 is 3600 = 2^{4} × 3^{2} × 5^{2}.

Thus, Total number of factors of 3600 is (4 + 1)(2 + 1)(2 + 1) = 5 × 3 × 3 = 45.

**Ques 4** : What is the number of factors of the number 504?

**Solution** : Prime Factorization of 504 is 504 = 2^{3} × 3^{2} × 7^{1}.

Thus, Total number of factors of 504 is (3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24.

**Ques 5** : Find the total number of factors of 180.

**Solution** : Prime Factorization of 180 is 180 = 2^{2} × 3^{2} × 5^{1}.

Thus, Total number of factors of 180 is (2 + 1)(2 + 1)(1 + 1) = 3 × 3 × 2 = 18.

**Ques 6** : What is the number of factors of 6480?

**Solution** : Prime Factorization of 6480 is 6480 = 2^{4} x 3^{4} x 5^{1}.

Thus, Total number of factors of 6480 is (4 + 1)(4 + 1)(1 + 1) = 5 × 5 × 2 = 50.

**Ques 7** : What is the number of factors of 420?

**Solution** : Prime Factorization of 420 is 420 = 2^{2} x 3^{1} x 5^{1 }x 7^{1}.

Thus, Total number of factors of 420 is (2 + 1)(1 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 × 2 = 24.

**Read** : Get Factors and Multiples of 1 to 100

**Read** : How to Find Number of EVEN Factors of Any Number

**Read** : How to Find Number of ODD Factors of Any Number

**Read** : How To Find Sum of Factors of Any Number

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]]>The post Factors and Multiples Tables | All | Even | Odd | Prime | Sum | Product of Factors appeared first on .

]]>Hi Students, welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * Factors and Multiples of 1 to 100*. All of these factors and multiples of 1 to 100 are given in table below. Apart from these, you will get all the factors, even factors, odd factors, prime factors, sum of factors and product of factors. And, also you will get first 10 multiples of numbers from 1 to 100.

**Read More About Factor Formula**

How To Find Number Of Factor of Any Number

How To Find Number of Even Factor of Any Number

How To Find Number of Odd Factor of Any Number

How To Find Sum of Factors of Any Number

How to Find Product of Factors of Any Number

How To Find Number of Factor Pairs of Any Number

Factors of 1 | |||||||
---|---|---|---|---|---|---|---|

Factors of 1 |
1 | ||||||

Number of Factors of 1 |
1 | ||||||

Even Factors of 1 |
None | ||||||

Odd Factors of 1 |
1 | ||||||

Prime Factors of 1 |
None | ||||||

Sum of Factors of 1 |
1 | ||||||

Product of Factors of 1 |
1 | ||||||

Multiples of 1 |
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 | ||||||

Factors of 2 | |||||||

Factors of 2 |
1, 2 | ||||||

Number of Factors of 2 |
2 | ||||||

Even Factors of 2 |
2 | ||||||

Odd Factors of 2 |
1 | ||||||

Prime Factors of 2 |
2 | ||||||

Sum of Factors of 2 |
3 | ||||||

Product of Factors of 2 |
2 | ||||||

Multiples of 2 |
2, 4, 6, 8, 10, 12, 14, 16, 18, 20 | ||||||

Factors of 3 | |||||||

Factors of 3 |
1, 3 | ||||||

Number of Factors of 3 |
2 | ||||||

Even Factors of 3 |
None | ||||||

Odd Factors of 3 |
1, 3 | ||||||

Prime Factors of 3 |
3 | ||||||

Sum of Factors of 3 |
4 | ||||||

Product of Factors of 3 |
3 | ||||||

Multiples of 3 |
3, 6, 9, 12, 15, 18, 21, 24, 27, 30 | ||||||

Factors of 4 | |||||||

Factors of 4 |
1, 2, 4 | ||||||

Number of Factors of 4 |
3 | ||||||

Even Factors of 4 |
2, 4 | ||||||

Odd Factors of 4 |
1 | ||||||

Prime Factors of 4 |
2 | ||||||

Sum of Factors of 4 |
7 | ||||||

Product of Factors of 4 |
8 | ||||||

Multiples of 4 |
4, 8, 12, 16, 20, 24, 28, 32, 36, 40 | ||||||

Factors of 5 | |||||||

Factors of 5 |
1, 5 | ||||||

Number of Factors of 5 |
2 | ||||||

Even Factors of 5 |
None | ||||||

Odd Factors of 5 |
1, 5 | ||||||

Prime Factors of 5 |
5 | ||||||

Sum of Factors of 5 |
6 | ||||||

Product of Factors of 5 |
5 | ||||||

Multiples of 5 |
5, 10, 15, 20, 25, 30, 35, 40, 45, 50 | ||||||

Factors of 6 | |||||||

Factors of 6 |
1, 2, 3, 6 | ||||||

Number of Factors of 6 |
4 | ||||||

Even Factors of 6 |
2, 6 | ||||||

Odd Factors of 6 |
1, 3 | ||||||

Prime Factors of 6 |
2, 3 | ||||||

Sum of Factors of 6 |
12 | ||||||

Product of Factors of 6 |
36 | ||||||

Multiples of 6 |
6, 12, 18, 24, 30, 36, 42, 48, 54, 60 | ||||||

Factors of 7 | |||||||

Factors of 7 |
1, 7 | ||||||

Number of Factors of 7 |
2 | ||||||

Even Factors of 7 |
None | ||||||

Odd Factors of 7 |
1, 7 | ||||||

Prime Factors of 7 |
7 | ||||||

Sum of Factors of 7 |
8 | ||||||

Product of Factors of 7 |
7 | ||||||

Multiples of 7 |
7, 14, 21, 28, 35, 42, 49, 56, 63, 70 | ||||||

Factors of 8 | |||||||

Factors of 8 |
1, 2, 4, 8 | ||||||

Number of Factors of 8 |
4 | ||||||

Even Factors of 8 |
2, 4, 8 | ||||||

Odd Factors of 8 |
1 | ||||||

Prime Factors of 8 |
2 | ||||||

Sum of Factors of 8 |
15 | ||||||

Product of Factors of 8 |
64 | ||||||

Multiples of 8 |
8, 16, 24, 32, 40, 48, 56, 64, 72, 80 | ||||||

Factors of 9 | |||||||

Factors of 9 |
1, 3, 9 | ||||||

Number of Factors of 9 |
3 | ||||||

Even Factors of 9 |
None | ||||||

Odd Factors of 9 |
1 | ||||||

Prime Factors of 9 |
3 | ||||||

Sum of Factors of 9 |
13 | ||||||

Product of Factors of 9 |
27 | ||||||

Multiples of 9 |
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 | ||||||

Factors of 10 | |||||||

Factors of 10 |
1, 2, 5, 10 | ||||||

Number of Factors of 10 |
4 | ||||||

Even Factors of 10 |
2, 10 | ||||||

Odd Factors of 10 |
1 | ||||||

Prime Factors of 10 |
2, 5 | ||||||

Sum of Factors of 10 |
18 | ||||||

Product of Factors of 10 |
100 | ||||||

Multiples of 10 |
10, 20, 30, 40, 50, 60, 70, 80, 90, 100 | ||||||

Factors of 11 | |||||||

Factors of 11 |
1, 11 | ||||||

Number of Factors of 11 |
2 | ||||||

Even Factors of 11 |
None | ||||||

Odd Factors of 11 |
1, 11 | ||||||

Prime Factors of 11 |
11 | ||||||

Sum of Factors of 11 |
12 | ||||||

Product of Factors of 1 |
11 | ||||||

Multiples of 11 |
11, 22, 33, 44, 55, 66, 77, 88, 99, 110 | ||||||

Factors of 12 | |||||||

Factors of 12 |
1, 2, 3, 4, 6, 12 | ||||||

Number of Factors of 12 |
6 | ||||||

Even Factors of 12 |
2, 4, 6, 12 | ||||||

Odd Factors of 12 |
1, 3 | ||||||

Prime Factors of 12 |
2, 3 | ||||||

Sum of Factors of 12 |
28 | ||||||

Product of Factors of 12 |
1728 | ||||||

Multiples of 12 |
12, 24, 36, 48, 60, 72, 84, 96, 108, 120 | ||||||

Factors of 13 | |||||||

Factors of 13 |
1, 13 | ||||||

Number of Factors of 13 |
2 | ||||||

Even Factors of 13 |
None | ||||||

Odd Factors of 13 |
1, 13 | ||||||

Prime Factors of 13 |
13 | ||||||

Sum of Factors of 13 |
14 | ||||||

Product of Factors of 13 |
13 | ||||||

Multiples of 13 |
13, 26, 39, 52, 65, 78, 91, 104, 117, 130 | ||||||

Factors of 14 | |||||||

Factors of 14 |
1, 2, 7, 14 | ||||||

Number of Factors of 14 |
4 | ||||||

Even Factors of 14 |
2, 14 | ||||||

Odd Factors of 14 |
1, 7 | ||||||

Prime Factors of 14 |
2, 7 | ||||||

Sum of Factors of 14 |
24 | ||||||

Product of Factors of 14 |
196 | ||||||

Multiples of 14 |
14, 28, 42, 56, 70, 84, 98, 112, 126, 140 | ||||||

Factors of 15 | |||||||

Factors of 15 |
1, 3, 5, 15 | ||||||

Number of Factors of 15 |
4 | ||||||

Even Factors of 15 |
None | ||||||

Odd Factors of 15 |
1, 3, 5, 15 | ||||||

Prime Factors of 15 |
3, 5 | ||||||

Sum of Factors of 15 |
24 | ||||||

Product of Factors of 15 |
225 | ||||||

Multiples of 15 |
15, 30, 45, 60, 75, 90, 105, 120, 135, 150 | ||||||

Factors of 16 | |||||||

Factors of 16 |
1, 2, 4, 8, 16 | ||||||

Number of Factors of 16 |
5 | ||||||

Even Factors of 16 |
2, 4, 8, 16 | ||||||

Odd Factors of 16 |
1 | ||||||

Prime Factors of 16 |
2 | ||||||

Sum of Factors of 16 |
31 | ||||||

Product of Factors of 16 |
1024 | ||||||

Multiples of 16 |
16, 32, 48, 64, 80, 96, 112, 128, 144, 160 | ||||||

Factors of 17 | |||||||

Factors of 17 |
1, 17 | ||||||

Number of Factors of 17 |
2 | ||||||

Even Factors of 17 |
None | ||||||

Odd Factors of 17 |
1, 17 | ||||||

Prime Factors of 17 |
17 | ||||||

Sum of Factors of 17 |
18 | ||||||

Product of Factors of 17 |
17 | ||||||

Multiples of 17 |
17, 34, 51, 68, 85, 102, 119, 136, 153, 170 | ||||||

Factors of 18 | |||||||

Factors of 18 |
1, 2, 3, 6, 9, 18 | ||||||

Number of Factors of 18 |
6, | ||||||

Even Factors of 18 |
2, 6, 18 | ||||||

Odd Factors of 18 |
1, 3, 9 | ||||||

Prime Factors of 18 |
2, 3 | ||||||

Sum of Factors of 18 |
39 | ||||||

Product of Factors of 18 |
5832 | ||||||

Multiples of 18 |
18, 36, 54, 72, 90, 108, 126, 144, 162, 180 | ||||||

Factors of 19 | |||||||

Factors of 19 |
1, 19 | ||||||

Number of Factors of 19 |
2 | ||||||

Even Factors of 19 |
None | ||||||

Odd Factors of 19 |
1, 19 | ||||||

Prime Factors of 19 |
19, | ||||||

Sum of Factors of 19 |
20 | ||||||

Product of Factors of 19 |
19 | ||||||

Multiples of 19 |
19, 38, 57, 76, 95, 114, 133, 152, 171, 190 | ||||||

Factors of 20 | |||||||

Factors of 20 |
1, 2, 4, 5, 10, 20 | ||||||

Number of Factors of 20 |
6 | ||||||

Even Factors of 20 |
2, 4, 10, 20 | ||||||

Odd Factors of 20 |
1, 5 | ||||||

Prime Factors of 20 |
2, 5 | ||||||

Sum of Factors of 20 |
42 | ||||||

Product of Factors of 20 |
8000 | ||||||

Multiples of 20 |
20, 40, 60, 80, 100, 120, 140, 160, 180, 200 | ||||||

Factors of 21 | |||||||

Factors of 21 |
1, 3, 7, 21 | ||||||

Number of Factors of 21 |
4 | ||||||

Even Factors of 21 |
None | ||||||

Odd Factors of 21 |
1, 3, 7, 21 | ||||||

Prime Factors of 21 |
3, 7 | ||||||

Sum of Factors of 21 |
32 | ||||||

Product of Factors of 21 |
144 | ||||||

Multiples of 21 |
21, 42, 63, 84, 105, 126, 147, 168, 189, 210 | ||||||

Factors of 22 | |||||||

Factors of 22 |
1, 2, 11, 22 | ||||||

Number of Factors of 22 |
4 | ||||||

Even Factors of 22 |
2, 22 | ||||||

Odd Factors of 22 |
1, 11 | ||||||

Prime Factors of 22 |
2, 11 | ||||||

Sum of Factors of 22 |
36 | ||||||

Product of Factors of 22 |
484 | ||||||

Multiples of 22 |
22, 44, 66, 88, 110, 132, 154, 176, 198, 220 | ||||||

Factors of 23 | |||||||

Factors of 23 |
1, 23 | ||||||

Number of Factors of 23 |
2 | ||||||

Even Factors of 23 |
None | ||||||

Odd Factors of 23 |
1, 23 | ||||||

Prime Factors of 23 |
23 | ||||||

Sum of Factors of 23 |
24 | ||||||

Product of Factors of 23 |
23 | ||||||

Multiples of 23 |
23, 46, 69, 92, 115, 138, 161, 184, 207, 230 | ||||||

Factors of 24 | |||||||

Factors of 24 |
1, 2, 3, 4, 6, 8, 12, 24 | ||||||

Number of Factors of 24 |
8 | ||||||

Even Factors of 24 |
2, 4, 6, 8, 12, 24 | ||||||

Odd Factors of 24 |
1, 3 | ||||||

Prime Factors of 24 |
2, 3 | ||||||

Sum of Factors of 24 |
60 | ||||||

Product of Factors of 24 |
331776 | ||||||

Multiples of 24 |
24, 48, 72, 96, 120, 144, 168, 192, 216, 240 | ||||||

Factors of 25 | |||||||

Factors of 25 |
1, 5, 25 | ||||||

Number of Factors of 25 |
3 | ||||||

Even Factors of 25 |
None | ||||||

Odd Factors of 25 |
1, 5, 25 | ||||||

Prime Factors of 25 |
5 | ||||||

Sum of Factors of 25 |
31 | ||||||

Product of Factors of 25 |
125 | ||||||

Multiples of 25 |
25, 50, 75, 100, 125, 150, 175, 200, 225, 250 | ||||||

Factors of 26 | |||||||

Factors of 26 |
1, 2, 13, 26 | ||||||

Number of Factors of 26 |
4 | ||||||

Even Factors of 26 |
2, 26 | ||||||

Odd Factors of 26 |
1, 13 | ||||||

Prime Factors of 26 |
2, 13 | ||||||

Sum of Factors of 26 |
42 | ||||||

Product of Factors of 26 |
676 | ||||||

Multiples of 26 |
26, 52, 78, 104, 130, 156, 182, 208, 234, 260 | ||||||

Factors of 27 | |||||||

Factors of 27 |
1, 3, 9, 27 | ||||||

Number of Factors of 27 |
4 | ||||||

Even Factors of 27 |
None | ||||||

Odd Factors of 27 |
1, 3, 9, 27 | ||||||

Prime Factors of 27 |
3 | ||||||

Sum of Factors of 27 |
40 | ||||||

Product of Factors of 27 |
729 | ||||||

Multiples of 27 |
27, 54, 81, 108, 135, 162, 198, 216, 243, 270 | ||||||

Factors of 28 | |||||||

Factors of 28 |
1, 2, 4, 7, 14, 28 | ||||||

Number of Factors of 28 |
6 | ||||||

Even Factors of 28 |
2, 4, 14, 28 | ||||||

Odd Factors of 28 |
1, 7 | ||||||

Prime Factors of 28 |
2, 7 | ||||||

Sum of Factors of 28 |
56 | ||||||

Product of Factors of 22 |
21952 | ||||||

Multiples of 28 |
28, 56, 84, 112, 140, 168, 196, 224, 252, 280 | ||||||

Factors of 29 | |||||||

Factors of 29 |
1, 29 | ||||||

Number of Factors of 29 |
2 | ||||||

Even Factors of 29 |
None | ||||||

Odd Factors of 29 |
1, 29 | ||||||

Prime Factors of 29 |
29 | ||||||

Sum of Factors of 29 |
30 | ||||||

Product of Factors of 29 |
29 | ||||||

Multiples of 29 |
29, 58, 87, 116, 145, 174, 203, 232, 261, 290 | ||||||

Factors of 30 | |||||||

Factors of 30 |
1, 2, 3, 5, 6, 10, 15, 30 | ||||||

Number of Factors of 30 |
8 | ||||||

Even Factors of 30 |
2, 6, 10, 30 | ||||||

Odd Factors of 30 |
1, 3, 5, 15 | ||||||

Prime Factors of 30 |
2, 3, 5 | ||||||

Sum of Factors of 30 |
72 | ||||||

Product of Factors of 30 |
810000 | ||||||

Multiples of 30 |
30, 60, 90, 120, 150, 180, 210, 240, 270, 300 | ||||||

Factors of 31 | |||||||

Factors of 31 |
1, 31 | ||||||

Number of Factors of 31 |
2 | ||||||

Even Factors of 31 |
None | ||||||

Odd Factors of 31 |
1, 31 | ||||||

Prime Factors of 31 |
31 | ||||||

Sum of Factors of 31 |
32 | ||||||

Product of Factors of 31 |
31 | ||||||

Multiples of 31 |
31, 62, 93, 124, 155, 186, 217, 248, 279, 310 | ||||||

Factors of 32 | |||||||

Factors of 32 |
1, 2, 4, 8, 16, 32 | ||||||

Number of Factors of 3 |
6 | ||||||

Even Factors of 3 |
2, 4, 8, 16, 32 | ||||||

Odd Factors of 3 |
1 | ||||||

Prime Factors of 3 |
2 | ||||||

Sum of Factors of 3 |
33 | ||||||

Product of Factors of 3 |
32768 | ||||||

Multiples of 3 |
32, 64, 96, 128, 160, 192, 224, 256, 288, 320 | ||||||

Factors of 33 | |||||||

Factors of 33 |
1, 3, 11, 33 | ||||||

Number of Factors of 33 |
4 | ||||||

Even Factors of 33 |
None | ||||||

Odd Factors of 33 |
1, 3, 11, 33 | ||||||

Prime Factors of 33 |
3, 11 | ||||||

Sum of Factors of 33 |
48 | ||||||

Product of Factors of 33 |
1089 | ||||||

Multiples of 33 |
33, 66, 99, 132, 165, 198, 231, 264, 297, 330 | ||||||

Factors of 34 | |||||||

Factors of 34 |
1, 2, 17, 34 | ||||||

Number of Factors of 34 |
4 | ||||||

Even Factors of 34 |
2, 34 | ||||||

Odd Factors of 34 |
1, 17 | ||||||

Prime Factors of 34 |
2, 17 | ||||||

Sum of Factors of 34 |
54 | ||||||

Product of Factors of 34 |
1156 | ||||||

Multiples of 34 |
34, 68, 102, 136, 170, 204, 238, 272, 306, 340 | ||||||

Factors of 35 | |||||||

Factors of 35 |
1, 5, 7, 35 | ||||||

Number of Factors of 35 |
4 | ||||||

Even Factors of 35 |
None | ||||||

Odd Factors of 35 |
1, 5, 7, 35 | ||||||

Prime Factors of 35 |
5, 7 | ||||||

Sum of Factors of 35 |
48 | ||||||

Product of Factors of 35 |
1225 | ||||||

Multiples of 35 |
35, 70, 105, 140, 175, 210, 245, 280, 315, 350 | ||||||

Factors of 36 | |||||||

Factors of 36 |
1, 2, 3, 4, 6, 9, 12, 18, 36 | ||||||

Number of Factors of 36 |
9 | ||||||

Even Factors of 36 |
2, 4, 6, 12, 18, 36 | ||||||

Odd Factors of 36 |
1, 3, 9 | ||||||

Prime Factors of 36 |
2, 3 | ||||||

Sum of Factors of 36 |
91 | ||||||

Product of Factors of 36 |
10077696 | ||||||

Multiples of 36 |
36, 72, 108, 144, 180, 216, 252, 288, 324, 360 | ||||||

Factors of 37 | |||||||

Factors of 37 |
1, 37 | ||||||

Number of Factors of 37 |
2 | ||||||

Even Factors of 37 |
None | ||||||

Odd Factors of 37 |
1, 37 | ||||||

Prime Factors of 37 |
37 | ||||||

Sum of Factors of 37 |
38 | ||||||

Product of Factors of 37 |
37 | ||||||

Multiples of 37 |
37, 74, 111, 148, 185, 222, 259, 296, 333, 370 | ||||||

Factors of 38 | |||||||

Factors of 38 |
1, 2, 19, 38 | ||||||

Number of Factors of 38 |
4 | ||||||

Even Factors of 38 |
2, 38 | ||||||

Odd Factors of 38 |
1, 19 | ||||||

Prime Factors of 38 |
2, 19 | ||||||

Sum of Factors of 38 |
60 | ||||||

Product of Factors of 38 |
1444 | ||||||

Multiples of 38 |
38, 76, 114, 152, 190, 228, 266, 304, 342, 380 | ||||||

Factors of 39 | |||||||

Factors of 39 |
1, 3, 13, 39 | ||||||

Number of Factors of 39 |
4 | ||||||

Even Factors of 39 |
None | ||||||

Odd Factors of 39 |
1, 3, 13, 39 | ||||||

Prime Factors of 39 |
3, 13 | ||||||

Sum of Factors of 39 |
56 | ||||||

Product of Factors of 39 |
1521 | ||||||

Multiples of 39 |
39, 78, 117, 156, 195, 234, 273, 312, 351, 390 | ||||||

Factors of 40 | |||||||

Factors of 40 |
1, 2, 4, 5, 8, 10, 20, 40 | ||||||

Number of Factors of 40 |
8 | ||||||

Even Factors of 40 |
2, 4, 8, 10, 20, 40 | ||||||

Odd Factors of 40 |
1, 5 | ||||||

Prime Factors of 40 |
2, 5 | ||||||

Sum of Factors of 40 |
90 | ||||||

Product of Factors of 40 |
160000 | ||||||

Multiples of 40 |
40, 80, 120, 160, 200, 240, 280, 320, 360, 400 | ||||||

Factors of 41 | |||||||

Factors of 41 |
1, 41 | ||||||

Number of Factors of 41 |
2 | ||||||

Even Factors of 41 |
None | ||||||

Odd Factors of 41 |
1, 41 | ||||||

Prime Factors of 41 |
41 | ||||||

Sum of Factors of 41 |
45 | ||||||

Product of Factors of 41 |
41 | ||||||

Multiples of 41 |
41, 82, 123, 164, 205, 246, 287, 328, 369, 410 | ||||||

Factors of 42 | |||||||

Factors of 42 |
1, 2, 3, 6, 7, 14, 21, 42 | ||||||

Number of Factors of 42 |
8 | ||||||

Even Factors of 42 |
2, 6, 14, 42 | ||||||

Odd Factors of 42 |
1, 3, 7, 21 | ||||||

Prime Factors of 42 |
2, 3, 7 | ||||||

Sum of Factors of 42 |
96 | ||||||

Product of Factors of 42 |
3111696 | ||||||

Multiples of 42 |
42, 84, 126, 168, 210, 252, 294, 336, 378, 420 | ||||||

Factors of 43 | |||||||

Factors of 43 |
1, 43 | ||||||

Number of Factors of 43 |
2 | ||||||

Even Factors of 43 |
None | ||||||

Odd Factors of 43 |
1, 43 | ||||||

Prime Factors of 43 |
43 | ||||||

Sum of Factors of 43 |
44 | ||||||

Product of Factors of 43 |
43 | ||||||

Multiples of 43 |
43, 86, 129, 172, 215, 258, 301, 344, 387, 430 | ||||||

Factors of 44 | |||||||

Factors of 44 |
1, 2, 4, 11, 22, 44 | ||||||

Number of Factors of 44 |
6 | ||||||

Even Factors of 44 |
2, 4, 22, 44 | ||||||

Odd Factors of 44 |
1, 11 | ||||||

Prime Factors of 44 |
2, 11 | ||||||

Sum of Factors of 44 |
84 | ||||||

Product of Factors of 44 |
85184 | ||||||

Multiples of 44 |
44, 88, 132, 176, 220, 264, 308, 352, 396 440 | ||||||

Factors of 45 | |||||||

Factors of 45 |
1, 3, 5, 9, 15, 45 | ||||||

Number of Factors of 45 |
6 | ||||||

Even Factors of 45 |
None | ||||||

Odd Factors of 45 |
1, 3, 5, 9, 15, 45 | ||||||

Prime Factors of 45 |
3, 5 | ||||||

Sum of Factors of 45 |
78 | ||||||

Product of Factors of 45 |
91125 | ||||||

Multiples of 45 |
45, 90, 135, 180, 225, 270, 315, 360, 405, 450 | ||||||

Factors of 46 | |||||||

Factors of 46 |
1, 2, 23, 46 | ||||||

Number of Factors of 46 |
4 | ||||||

Even Factors of 46 |
2, 46 | ||||||

Odd Factors of 46 |
1, 23 | ||||||

Prime Factors of 46 |
2, 23 | ||||||

Sum of Factors of 46 |
72 | ||||||

Product of Factors of 46 |
2116 | ||||||

Multiples of 46 |
46, 92, 138, 184, 230, 276, 322, 368, 414, 460 | ||||||

Factors of 47 | |||||||

Factors of 47 |
1, 47 | ||||||

Number of Factors of 47 |
2 | ||||||

Even Factors of 47 |
None | ||||||

Odd Factors of 47 |
1, 47 | ||||||

Prime Factors of 47 |
47 | ||||||

Sum of Factors of 47 |
48 | ||||||

Product of Factors of 47 |
47 | ||||||

Multiples of 47 |
47, 94, 141, 188, 235, 282, 329, 376, 423, 470 | ||||||

Factors of 48 | |||||||

Factors of 48 |
1, 2, 3, 4, 6, 8, 12, 16, 24, 48 | ||||||

Number of Factors of 48 |
10 | ||||||

Even Factors of 48 |
2, 4, 6, 8, 12, 16, 24, 48 | ||||||

Odd Factors of 48 |
1, 3 | ||||||

Prime Factors of 48 |
2, 3 | ||||||

Sum of Factors of 48 |
124 | ||||||

Product of Factors of 48 |
254803968 | ||||||

Multiples of 48 |
48, 96, 144, 192, 240, 288, 336, 384, 432, 480 | ||||||

Factors of 49 | |||||||

Factors of 49 |
1, 7, 49 | ||||||

Number of Factors of 49 |
3 | ||||||

Even Factors of 49 |
None | ||||||

Odd Factors of 49 |
1, 7, 49 | ||||||

Prime Factors of 49 |
7 | ||||||

Sum of Factors of 49 |
57 | ||||||

Product of Factors of 49 |
343 | ||||||

Multiples of 49 |
49, 38, 147, 196, 245, 294, 343, 392, 441, 490 | ||||||

Factors of 50 | |||||||

Factors of 50 |
1, 2, 5, 10, 25, 50 | ||||||

Number of Factors of 50 |
6 | ||||||

Even Factors of 50 |
2, 10, 50 | ||||||

Odd Factors of 50 |
1, 5, 25 | ||||||

Prime Factors of 50 |
2, 5 | ||||||

Sum of Factors of 50 |
93 | ||||||

Product of Factors of 50 |
125000 | ||||||

Multiples of 50 |
50, 100, 150, 200, 250, 300, 350, 400, 450, 500 | ||||||

Factors of 51 | |||||||

Factors of 51 |
1, 3, 17, 51 | ||||||

Number of Factors of 51 |
2 | ||||||

Even Factors of 51 |
None | ||||||

Odd Factors of 51 |
1, 3, 17, 51 | ||||||

Prime Factors of 51 |
3, 17 | ||||||

Sum of Factors of 51 |
72 | ||||||

Product of Factors of 51 |
2601 | ||||||

Multiples of 51 |
51, 102, 153, 204, 255, 306, 357, 408, 459, 510 | ||||||

Factors of 52 | |||||||

Factors of 52 |
1, 2, 4, 13, 26, 52 | ||||||

Number of Factors of 52 |
6 | ||||||

Even Factors of 52 |
2, 4, 26, 52 | ||||||

Odd Factors of 52 |
1, 13 | ||||||

Prime Factors of 52 |
2, 13 | ||||||

Sum of Factors of 52 |
98 | ||||||

Product of Factors of 52 |
140608 | ||||||

Multiples of 52 |
52, 104, 156, 208, 260, 312, 364, 416, 468, 520 | ||||||

Factors of 53 | |||||||

Factors of 53 |
1, 53 | ||||||

Number of Factors of 53 |
2 | ||||||

Even Factors of 53 |
None | ||||||

Odd Factors of 53 |
1, 53 | ||||||

Prime Factors of 53 |
53 | ||||||

Sum of Factors of 53 |
54 | ||||||

Product of Factors of 53 |
53 | ||||||

Multiples of 53 |
53, 106, 159, 212, 265, 318, 371, 424, 477, 530 | ||||||

Factors of 54 | |||||||

Factors of 54 |
1, 2, 3, 6, 9, 18, 27, 54 | ||||||

Number of Factors of 54 |
8 | ||||||

Even Factors of 54 |
2, 6, 18, 54 | ||||||

Odd Factors of 54 |
1, 3, 9, 27 | ||||||

Prime Factors of 54 |
2, 3 | ||||||

Sum of Factors of 54 |
120 | ||||||

Product of Factors of 54 |
8503056 | ||||||

Multiples of 54 |
54, 108, 162, 216, 270, 324, 378, 432, 486, 540 | ||||||

Factors of 55 | |||||||

Factors of 55 |
1, 5, 11, 55 | ||||||

Number of Factors of 55 |
4 | ||||||

Even Factors of 55 |
None | ||||||

Odd Factors of 55 |
1, 5, 11, 55 | ||||||

Prime Factors of 55 |
5, 11 | ||||||

Sum of Factors of 55 |
72 | ||||||

Product of Factors of 55 |
3025 | ||||||

Multiples of 55 |
55, 110, 165, 220, 275, 330, 385, 440, 495, 550 | ||||||

Factors of 56 | |||||||

Factors of 56 |
1, 2, 4, 7, 8, 14, 28, 56 | ||||||

Number of Factors of 56 |
8 | ||||||

Even Factors of 56 |
2, 4, 8, 14, 28, 56 | ||||||

Odd Factors of 56 |
1, 7 | ||||||

Prime Factors of 56 |
2, 7 | ||||||

Sum of Factors of 56 |
120 | ||||||

Product of Factors of 56 |
9834496 | ||||||

Multiples of 56 |
56, 112, 168, 224, 280, 336, 392, 448, 504, 560 | ||||||

Factors of 57 | |||||||

Factors of 57 |
1, 3, 19, 57 | ||||||

Number of Factors of 57 |
4 | ||||||

Even Factors of 57 |
None | ||||||

Odd Factors of 57 |
1, 3, 19, 57 | ||||||

Prime Factors of 57 |
3, 19 | ||||||

Sum of Factors of 57 |
80 | ||||||

Product of Factors of 57 |
3249 | ||||||

Multiples of 57 |
57, 114, 171, 228, 285, 342, 399, 456, 513, 570 | ||||||

Factors of 58 | |||||||

Factors of 58 |
1, 2, 29, 58 | ||||||

Number of Factors of 58 |
4 | ||||||

Even Factors of 58 |
2, 58 | ||||||

Odd Factors of 58 |
1, 29 | ||||||

Prime Factors of 58 |
2, 29 | ||||||

Sum of Factors of 58 |
90 | ||||||

Product of Factors of 58 |
3364 | ||||||

Multiples of 58 |
58, 116, 174, 232, 290, 348, 406, 464, 522, 580 | ||||||

Factors of 59 | |||||||

Factors of 59 |
1, 59 | ||||||

Number of Factors of 59 |
2 | ||||||

Even Factors of 59 |
None | ||||||

Odd Factors of 59 |
1, 59 | ||||||

Prime Factors of 59 |
59 | ||||||

Sum of Factors of 59 |
60 | ||||||

Product of Factors of 59 |
59 | ||||||

Multiples of 59 |
59, 118, 177, 236, 295, 354, 413, 472, 531, 590 | ||||||

Factors of 60 | |||||||

Factors of 60 |
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 | ||||||

Number of Factors of 60 |
12 | ||||||

Even Factors of 60 |
2, 4, 6, 10, 12, 20, 30, 60 | ||||||

Odd Factors of 60 |
1, 3, 5, 15 | ||||||

Prime Factors of 60 |
2, 3, 5 | ||||||

Sum of Factors of 60 |
168 | ||||||

Product of Factors of 60 |
46656000000 | ||||||

Multiples of 60 |
60, 120, 180, 240, 300, 360, 420, 480, 540, 600 | ||||||

Factors of 6 | |||||||

Factors of 61 |
1, 61 | ||||||

Number of Factors of 61 |
2 | ||||||

Even Factors of 61 |
None | ||||||

Odd Factors of 61 |
1, 61 | ||||||

Prime Factors of 61 |
61 | ||||||

Sum of Factors of 61 |
62 | ||||||

Product of Factors of 61 |
61 | ||||||

Multiples of 61 |
61, 122, 183, 244, 305, 366, 427, 488, 549, 610 | ||||||

Factors of 62 | |||||||

Factors of 62 |
1, 2, 31, 62 | ||||||

Number of Factors of 62 |
4 | ||||||

Even Factors of 62 |
2, 62 | ||||||

Odd Factors of 62 |
1, 31 | ||||||

Prime Factors of 62 |
2, 31 | ||||||

Sum of Factors of 62 |
96 | ||||||

Product of Factors of 62 |
3844 | ||||||

Multiples of 62 |
62, 124, 186, 248, 310, 372, 434, 496, 558, 620 | ||||||

Factors of 63 | |||||||

Factors of 63 |
1, 3, 7, 9, 21, 63 | ||||||

Number of Factors of 63 |
6 | ||||||

Even Factors of 63 |
None | ||||||

Odd Factors of 63 |
1, 3, 7, 9, 21, 63 | ||||||

Prime Factors of 63 |
3, 7 | ||||||

Sum of Factors of 63 |
104 | ||||||

Product of Factors of 63 |
250047 | ||||||

Multiples of 63 |
63, 126, 189, 252, 315, 378, 441, 504, 567, 630 | ||||||

Factors of 64 | |||||||

Factors of 64 |
1, 2, 4, 8, 16, 32, 64 | ||||||

Number of Factors of 64 |
7 | ||||||

Even Factors of 64 |
2, 4, 8, 16, 32, 64 | ||||||

Odd Factors of 64 |
1 | ||||||

Prime Factors of 64 |
2 | ||||||

Sum of Factors of 64 |
127 | ||||||

Product of Factors of 64 |
2097152 | ||||||

Multiples of 64 |
64, 128, 192, 320, 384, 448, 512, 576, 640 | ||||||

Factors of 65 | |||||||

Factors of 65 |
1, 5, 13, 65 | ||||||

Number of Factors of 65 |
4 | ||||||

Even Factors of 65 |
None | ||||||

Odd Factors of 65 |
1, 5, 13, 65 | ||||||

Prime Factors of 65 |
5, 13 | ||||||

Sum of Factors of 65 |
84 | ||||||

Product of Factors of 65 |
4225 | ||||||

Multiples of 65 |
65, 130, 195, 260, 325, 390, 455, 520, 585, 650 | ||||||

Factors of 66 | |||||||

Factors of 66 |
1, 2, 3, 6, 11, 22, 33, 66 | ||||||

Number of Factors of 66 |
8 | ||||||

Even Factors of 66 |
2, 6, 22, 66 | ||||||

Odd Factors of 66 |
1, 3, 11, 33 | ||||||

Prime Factors of 66 |
2, 3, 11 | ||||||

Sum of Factors of 66 |
144 | ||||||

Product of Factors of 66 |
18974736 | ||||||

Multiples of 66 |
66, 132, 198, 264, 330, 396, 462, 528, 594, 660 | ||||||

Factors of 67 | |||||||

Factors of 67 |
1, 67 | ||||||

Number of Factors of 67 |
2 | ||||||

Even Factors of 67 |
None | ||||||

Odd Factors of 67 |
1, 67 | ||||||

Prime Factors of 67 |
67 | ||||||

Sum of Factors of 67 |
68 | ||||||

Product of Factors of 67 |
67 | ||||||

Multiples of 67 |
67, 134, 201, 268, 335, 402, 469, 536, 603, 670 | ||||||

Factors of 68 | |||||||

Factors of 68 |
1, 2, 4, 17, 34, 68 | ||||||

Number of Factors of 68 |
6 | ||||||

Even Factors of 68 |
2, 4, 34, 68 | ||||||

Odd Factors of 68 |
1, 17 | ||||||

Prime Factors of 68 |
2, 17 | ||||||

Sum of Factors of 68 |
126 | ||||||

Product of Factors of 68 |
314432 | ||||||

Multiples of 68 |
68, 136, 204, 272, 340, 408, 476, 544, 612, 680 | ||||||

Factors of 69 | |||||||

Factors of 69 |
1, 3, 23, 69 | ||||||

Number of Factors of 69 |
4 | ||||||

Even Factors of 69 |
None | ||||||

Odd Factors of 69 |
1, 3, 23, 69 | ||||||

Prime Factors of 69 |
3, 23 | ||||||

Sum of Factors of 69 |
96 | ||||||

Product of Factors of 69 |
4761 | ||||||

Multiples of 69 |
69, 138, 207, 276, 345, 414, 484, 552, 621, 690 | ||||||

Factors of 70 | |||||||

Factors of 70 |
1, 2, 5, 7, 10, 14, 35, 70 | ||||||

Number of Factors of 70 |
8 | ||||||

Even Factors of 70 |
2, 10, 14, 70 | ||||||

Odd Factors of 70 |
1, 5, 7, 35 | ||||||

Prime Factors of 70 |
2, 5, 7 | ||||||

Sum of Factors of 70 |
144 | ||||||

Product of Factors of 70 |
24010000 | ||||||

Multiples of 70 |
70, 140, 210, 280, 350, 420, 490, 560, 630, 700 | ||||||

Factors of 71 | |||||||

Factors of 71 |
1, 71 | ||||||

Number of Factors of 71 |
2 | ||||||

Even Factors of 71 |
None | ||||||

Odd Factors of 71 |
1, 71 | ||||||

Prime Factors of 71 |
71 | ||||||

Sum of Factors of 71 |
72 | ||||||

Product of Factors of 71 |
71 | ||||||

Multiples of 71 |
71, 142, 213, 284, 355, 426, 497, 568, 639, 700 | ||||||

Factors of 72 | |||||||

Factors of 72 |
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 | ||||||

Number of Factors of 72 |
11 | ||||||

Even Factors of 72 |
2, 4, 6, 8, 12, 18, 24, 36, 72 | ||||||

Odd Factors of 72 |
1, 3, 9 | ||||||

Prime Factors of 72 |
2, 3 | ||||||

Sum of Factors of 72 |
195 | ||||||

Product of Factors of 72 |
139314069504 | ||||||

Multiples of 72 |
72, 144, 216, 288, 360, 432, 504, 576, 648, 720 | ||||||

Factors of 73 | |||||||

Factors of 73 |
1, 73 | ||||||

Number of Factors of 73 |
2 | ||||||

Even Factors of 73 |
None | ||||||

Odd Factors of 73 |
1, 73 | ||||||

Prime Factors of 73 |
73 | ||||||

Sum of Factors of 73 |
74 | ||||||

Product of Factors of 73 |
73 | ||||||

Multiples of 73 |
73, 146, 219, 292, 365, 438, 511, 584, 657, 730 | ||||||

Factors of 74 | |||||||

Factors of 74 |
1, 2, 37, 74 | ||||||

Number of Factors of 74 |
4 | ||||||

Even Factors of 74 |
2, 74 | ||||||

Odd Factors of 74 |
1, 37 | ||||||

Prime Factors of 74 |
2, 37 | ||||||

Sum of Factors of 74 |
114 | ||||||

Product of Factors of 74 |
5476 | ||||||

Multiples of 74 |
74, 148, 222, 296, 370, 444, 518, 592, 666, 740 | ||||||

Factors of 75 | |||||||

Factors of 75 |
1, 3, 5, 15, 25, 75 | ||||||

Number of Factors of 75 |
6 | ||||||

Even Factors of 75 |
None | ||||||

Odd Factors of 75 |
1, 3, 5, 15, 25, 75 | ||||||

Prime Factors of 75 |
3, 5 | ||||||

Sum of Factors of 75 |
124 | ||||||

Product of Factors of 75 |
421875 | ||||||

Multiples of 75 |
75, 150, 225, 300, 375, 450, 525, 600, 675, 750 | ||||||

Factors of 76 | |||||||

Factors of 76 |
1, 2, 4, 19, 38, 76 | ||||||

Number of Factors of 76 |
6 | ||||||

Even Factors of 76 |
2, 4, 38, 76 | ||||||

Odd Factors of 76 |
1, 19 | ||||||

Prime Factors of 76 |
2, 19 | ||||||

Sum of Factors of 76 |
140 | ||||||

Product of Factors of 76 |
438976 | ||||||

Multiples of 76 |
76, 152, 228, 304, 380, 456, 532, 608, 684, 760 | ||||||

Factors of 77 | |||||||

Factors of 77 |
1, 7, 11, 77 | ||||||

Number of Factors of 77 |
4 | ||||||

Even Factors of 77 |
None | ||||||

Odd Factors of 77 |
1, 7, 11, 17 | ||||||

Prime Factors of 77 |
7, 11 | ||||||

Sum of Factors of 77 |
96 | ||||||

Product of Factors of 77 |
5929 | ||||||

Multiples of 77 |
77, 154, 231, 308, 385, 462, 539, 616, 693, 770 | ||||||

Factors of 78 | |||||||

Factors of 78 |
1, 2, 3, 6, 13, 26, 39, 78 | ||||||

Number of Factors of 78 |
8 | ||||||

Even Factors of 78 |
2, 6, 26, 78 | ||||||

Odd Factors of 78 |
1, 3, 13, 39 | ||||||

Prime Factors of 78 |
2, 3, 13 | ||||||

Sum of Factors of 78 |
168 | ||||||

Product of Factors of 78 |
37015056 | ||||||

Multiples of 78 |
78, 156, 234, 312, 390, 468, 546, 624, 702, 780 | ||||||

Factors of 79 | |||||||

Factors of 79 |
1, 79 | ||||||

Number of Factors of 79 |
2 | ||||||

Even Factors of 79 |
None | ||||||

Odd Factors of 79 |
1, 79 | ||||||

Prime Factors of 79 |
79 | ||||||

Sum of Factors of 79 |
80 | ||||||

Product of Factors of 79 |
79 | ||||||

Multiples of 79 |
79, 158, 237, 316, 395, 474, 553, 632, 711, 790 | ||||||

Factors of 80 | |||||||

Factors of 80 |
1, 2, 4, 5, 8, 10, 16, 20, 40, 80 | ||||||

Number of Factors of 80 |
10 | ||||||

Even Factors of 80 |
2, 4, 8, 10, 16, 20, 40, 80 | ||||||

Odd Factors of 80 |
1, 5 | ||||||

Prime Factors of 80 |
2, 5 | ||||||

Sum of Factors of 80 |
186 | ||||||

Product of Factors of 80 |
3276800000 | ||||||

Multiples of 80 |
80, 160, 240, 320, 400, 480, 560, 640, 720, 800 | ||||||

Factors of 81 | |||||||

Factors of 81 |
1, 3, 9, 27, 81 | ||||||

Number of Factors of 81 |
5 | ||||||

Even Factors of 81 |
None | ||||||

Odd Factors of 81 |
1, 3, 9, 27, 81 | ||||||

Prime Factors of 81 |
3 | ||||||

Sum of Factors of 81 |
121 | ||||||

Product of Factors of 81 |
59049 | ||||||

Multiples of 81 |
81, 162, 243, 324, 405, 486, 567, 648, 729, 810 | ||||||

Factors of 82 | |||||||

Factors of 82 |
1, 2, 41, 82 | ||||||

Number of Factors of 82 |
4 | ||||||

Even Factors of 82 |
2, 82 | ||||||

Odd Factors of 82 |
1, 41 | ||||||

Prime Factors of 82 |
2, 41 | ||||||

Sum of Factors of 82 |
126 | ||||||

Product of Factors of 82 |
6724 | ||||||

Multiples of 82 |
82, 164, 246, 328, 410, 492, 574, 656, 738, 820 | ||||||

Factors of 83 | |||||||

Factors of 83 |
1, 83 | ||||||

Number of Factors of 83 |
2 | ||||||

Even Factors of 83 |
None | ||||||

Odd Factors of 83 |
1, 83 | ||||||

Prime Factors of 83 |
83 | ||||||

Sum of Factors of 83 |
84 | ||||||

Product of Factors of 83 |
83 | ||||||

Multiples of 83 |
83, 166, 249, 332, 415, 498, 581, 664, 747, 830 | ||||||

Factors of 84 | |||||||

Factors of 84 |
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 | ||||||

Number of Factors of 84 |
12 | ||||||

Even Factors of 84 |
2, 4, 6, 12, 14, 28, 42, 84 | ||||||

Odd Factors of 84 |
1, 3, 7, 21 | ||||||

Prime Factors of 84 |
2, 3, 7 | ||||||

Sum of Factors of 84 |
224 | ||||||

Product of Factors of 84 |
351298031616 | ||||||

Multiples of 84 |
84, 168, 252, 336, 420, 504, 588, 672, 756, 840 | ||||||

Factors of 85 | |||||||

Factors of 85 |
1, 5, 17, 85 | ||||||

Number of Factors of 85 |
4 | ||||||

Even Factors of 85 |
None | ||||||

Odd Factors of 85 |
1, 5, 17, 85 | ||||||

Prime Factors of 85 |
5, 17 | ||||||

Sum of Factors of 85 |
108 | ||||||

Product of Factors of 85 |
7225 | ||||||

Multiples of 85 |
85, 170, 255, 340, 425, 510, 595, 680, 765, 850 | ||||||

Factors of 86 | |||||||

Factors of 86 |
1, 2, 43, 86 | ||||||

Number of Factors of 86 |
4 | ||||||

Even Factors of 86 |
2 | ||||||

Odd Factors of 86 |
1, 43, | ||||||

Prime Factors of 86 |
2, 43 | ||||||

Sum of Factors of 86 |
132 | ||||||

Product of Factors of 86 |
7396 | ||||||

Multiples of 86 |
86, 172, 258, 344, 430, 516, 602, 688, 774, 860 | ||||||

Factors of 87 | |||||||

Factors of 87 |
1, 3, 29, 87 | ||||||

Number of Factors of 87 |
4 | ||||||

Even Factors of 87 |
None | ||||||

Odd Factors of 87 |
1, 3, 29, 87 | ||||||

Prime Factors of 87 |
3, 29 | ||||||

Sum of Factors of 87 |
120 | ||||||

Product of Factors of 87 |
7569 | ||||||

Multiples of 87 |
87, 174, 261, 348, 435, 522, 609, 696, 783, 870 | ||||||

Factors of 88 | |||||||

Factors of 88 |
1, 2, 4, 8, 11, 22, 44, 88 | ||||||

Number of Factors of 88 |
8 | ||||||

Even Factors of 88 |
2, 4, 8, 22, 44, 88 | ||||||

Odd Factors of 88 |
1, 11 | ||||||

Prime Factors of 88 |
2, 11 | ||||||

Sum of Factors of 88 |
180 | ||||||

Product of Factors of 88 |
59969536 | ||||||

Multiples of 88 |
88, 176, 264, 352, 440, 528, 616, 704, 792, 880 | ||||||

Factors of 89 | |||||||

Factors of 89 |
1, 89 | ||||||

Number of Factors of 89 |
2 | ||||||

Even Factors of 89 |
None | ||||||

Odd Factors of 89 |
1, 89 | ||||||

Prime Factors of 89 |
89 | ||||||

Sum of Factors of 89 |
90 | ||||||

Product of Factors of 89 |
89 | ||||||

Multiples of 89 |
89, 178, 267, 356, 445, 534, 623, 712, 801, 890 | ||||||

Factors of 90 | |||||||

Factors of 90 |
1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 | ||||||

Number of Factors of 90 |
12 | ||||||

Even Factors of 90 |
2, 6, 10, 18, 30, 90 | ||||||

Odd Factors of 90 |
1, 3, 5, 9, 15, 45 | ||||||

Prime Factors of 90 |
2, 3, 5 | ||||||

Sum of Factors of 90 |
234 | ||||||

Product of Factors of 90 |
531441 | ||||||

Multiples of 90 |
90, 180, 270, 360, 450, 540, 630, 720, 810, 900 | ||||||

Factors of 91 | |||||||

Factors of 91 |
1, 7, 13, 91 | ||||||

Number of Factors of 91 |
4 | ||||||

Even Factors of 91 |
None | ||||||

Odd Factors of 91 |
1, 7, 13, 91 | ||||||

Prime Factors of 91 |
7, 13 | ||||||

Sum of Factors of 91 |
112 | ||||||

Product of Factors of 91 |
8281 | ||||||

Multiples of 91 |
91, 182, 273, 364, 455, 546, 637, 728, 819, 910 | ||||||

Factors of 92 | |||||||

Factors of 92 |
1, 2, 4, 23, 46, 92 | ||||||

Number of Factors of 92 |
6 | ||||||

Even Factors of 92 |
2, 4, 46, 92 | ||||||

Odd Factors of 92 |
1, 23 | ||||||

Prime Factors of 92 |
2, 23 | ||||||

Sum of Factors of 92 |
168 | ||||||

Product of Factors of 92 |
778688 | ||||||

Multiples of 92 |
92, 184, 276, 368, 460, 552, 644, 736, 828, 920 | ||||||

Factors of 93 | |||||||

Factors of 93 |
1, 3, 31, 93 | ||||||

Number of Factors of 93 |
4 | ||||||

Even Factors of 93 |
None | ||||||

Odd Factors of 93 |
1, 3, 31, 93 | ||||||

Prime Factors of 93 |
3, 31 | ||||||

Sum of Factors of 93 |
128 | ||||||

Product of Factors of 93 |
8649 | ||||||

Multiples of 93 |
93, 186, 279, 372, 465, 558, 651, 744, 837, 930 | ||||||

Factors of 94 | |||||||

Factors of 94 |
1, 2, 47, 94 | ||||||

Number of Factors of 94 |
4 | ||||||

Even Factors of 94 |
2, 94 | ||||||

Odd Factors of 94 |
1, 47 | ||||||

Prime Factors of 94 |
2, 47 | ||||||

Sum of Factors of 94 |
144 | ||||||

Product of Factors of 94 |
8836 | ||||||

Multiples of 94 |
94, 188, 282, 376, 470, 564, 658, 752, 846, 940 | ||||||

Factors of 95 | |||||||

Factors of 95 |
1, 5, 19, 95 | ||||||

Number of Factors of 95 |
4 | ||||||

Even Factors of 95 |
None | ||||||

Odd Factors of 95 |
1, 5, 19, 95 | ||||||

Prime Factors of 95 |
5, 19 | ||||||

Sum of Factors of 95 |
120 | ||||||

Product of Factors of 95 |
9025 | ||||||

Multiples of 95 |
95, 190, 285, 380, 475, 570, 665, 760, 855, 950 | ||||||

Factors of 96 | |||||||

Factors of 96 |
1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 | ||||||

Number of Factors of 96 |
12 | ||||||

Even Factors of 96 |
2, 4, 8, 12, 16, 24, 32, 48, 96 | ||||||

Odd Factors of 96 |
1, 3 | ||||||

Prime Factors of 96 |
2, 3 | ||||||

Sum of Factors of 96 |
252 | ||||||

Product of Factors of 96 |
782757789696 | ||||||

Multiples of 96 |
96, 192, 288, 384, 480, 576, 672, 768, 864, 960 | ||||||

Factors of 97 | |||||||

Factors of 97 |
1, 97 | ||||||

Number of Factors of 97 |
2 | ||||||

Even Factors of 97 |
None | ||||||

Odd Factors of 97 |
1, 97 | ||||||

Prime Factors of 97 |
97 | ||||||

Sum of Factors of 97 |
98, | ||||||

Product of Factors of 97 |
97 | ||||||

Multiples of 97 |
97, 194, 291, 388, 485, 582, 679, 776, 873, 970 | ||||||

Factors of 98 | |||||||

Factors of 98 |
1, 2, 7, 14, 49, 98 | ||||||

Number of Factors of 98 |
6 | ||||||

Even Factors of 98 |
2, 14, 98 | ||||||

Odd Factors of 98 |
1, 7, 49 | ||||||

Prime Factors of 98 |
2, 7 | ||||||

Sum of Factors of 98 |
171 | ||||||

Product of Factors of 98 |
941192 | ||||||

Multiples of 98 |
98, 196, 294, 392, 490, 588, 686, 784, 882, 980 | ||||||

Factors of 99 | |||||||

Factors of 99 |
1, 3, 9, 11, 33, 99 | ||||||

Number of Factors of 99 |
6 | ||||||

Even Factors of 99 |
None | ||||||

Odd Factors of 99 |
1, 3, 11, 33, 99 | ||||||

Prime Factors of 99 |
3, 11 | ||||||

Sum of Factors of 99 |
156 | ||||||

Product of Factors of 99 |
970299 | ||||||

Multiples of 99 |
99, 198, 297, 396, 495, 594, 693, 792, 891, 990 | ||||||

Factors of 100 | |||||||

Factors of 100 |
1, 2, 4, 5, 10, 20, 25, 50, 100 | ||||||

Number of Factors of 100 |
9 | ||||||

Even Factors of 100 |
2, 4, 10, 20, 50, 100 | ||||||

Odd Factors of 100 |
1, 5, 25 | ||||||

Prime Factors of 100 |
2, 5, | ||||||

Sum of Factors of 100 |
217 | ||||||

Product of Factors of 100 |
1000000000 | ||||||

Multiples of 100 |
100, 200, 300, 400, 500, 600, 700, 800, 900, 1000 |

**Read More About Factor Formula**

How To Find Number Of Factor of Any Number

How To Find Number of Even Factor of Any Number

How To Find Number of Odd Factor of Any Number

How To Find Sum of Factors of Any Number

How to Find Product of Factors of Any Number

How To Find Number of Factor Pairs of Any Number

Important points to remember in the above list.

The post Factors and Multiples Tables | All | Even | Odd | Prime | Sum | Product of Factors appeared first on .

]]>The post NMTC Stage 1 and 2 Sub Junior Level Previous Year Papers with Solutions appeared first on .

]]>Hi NMTC aspirants, **Amans Maths Blogs (AMB) **created an ebook of **NMTC Previous Year Papers with Solutions** according to class wise: for **Primary Level (Class 5 & 6)**, for **Sub-Junior Level (Class 7 & 8)** and **Junior Level (Class 9 & 10)**. In this post, you will get an ebook of **NMTC Stage 1 and 2 Sub Junior Level Previous Year Papers Solutions PDF**.

These books are the alternative and best books for NMTC preparation, because previous year questions papers with answer keys and solutions will give an idea of **NMTC syllabus** and their difficulty level of asked questions in * NMTC question paper* or

**NMTC** is a short form of National Mathematics Talent Competition. This is conducted by AMTI to identify the talent of the students in Mathematics.

Now, for more details about the **NMTC exam**, like its **NMTC eligibility**, medium of **NMTC papers**, examination fee, and **NMTC syllabus** etc, click here : **NMTC Notification**.

**AMTI** is a short form of Association Mathematics Teachers of India. It is an academic body of professionals and students who have a passion in Mathematics research and education. For the AMTI NMTC exam, you need amti previous question papers with solutions.

NMTC exams are in two stages. It is called as NMTC Stage 1 and Stage 2 or NMTC Screening Level and NMTC Final Level exam.

To clear **NMTC exam**, you need **NMTC Papers with Solutions **it will help you in the preparation of this exam, So I created an EBook of * NMTC Previous Question Papers with Solutions PDF*.

According to * AMTI*, the students who have an ability for unique creativity thinking a unique and imaginative considering, preparation to assault new and non-routine issues displaying a general numerical capacity proper to their level. they must start the preparation of

For the * NMTC preparation*, you need

Get EBook PDF of **NMTC Stage 1 and 2 Primary Level Papers with Solutions**

NMTC Sub Junior Level exam is known as Kaprekar Contest, its eligibility is for Class 7 and 8. Thus, to prepare for NMTC Sub Junior level exam, you need **NMTC Stage 1 Sub Junior Level Previous Years Questions Papers with Solutions**.

*Downloading Link of NMTC 2004 to 2019 Sub Junior Level Past Papers with Solutions is Given Below of this Post*

**Question 1 :**

A four-digit number of the form abaa is divisible by 33. The number of such four-digit number is

(A) 36

(B) 6

(C) 3

(D) 1

**Solution 1 :**

**Question 2 :**

If a + 2a + 3a + … + 1000a = 2b + 4b + 6b + … + 2000b = 3c + 6c + 9c + … + 3000c, then a : b : c is

(A) 1 : 2 : 3

(B) 3 : 2 : 1

(C) 2 : 3 : 6

(D) 6 : 3 : 2

**Solution 2 :**

*Downloading Link of NMTC 2004 to 2019 Sub Junior Level Past Papers with Solutions is Given Below of this Post*

**Question 1 :**

You join a job. Your pay for the first day is Rs 5. Each day after that your pay will be twice as much as it was the day before. Your pay on the tenth will be

(A) Rs. 100

(B) Rs. 250

(C) Rs. 5120

(D) Rs. 2560

**Solution 1 :**

**Question 2 :**

Using all the digits 1 to 9 only one, how many nine digit prime numbers can you write?

(A) 1

(B) 0

(C) 9

(D) More than 100

**Solution 2 :**

*Downloading Link of NMTC 2004 to 2019 Sub Junior Level Past Papers with Solutions is Given Below of this Post*

**Question 1 :**

Given the alphabetic where each letter represents a different digit. The value of C in the least value for ABCD is

(A) 1

(B) 2

(C) 6

(D) 7

**Solution 1 :**

**Question 2 :**

If 36a^{4} = a^{6}, then the value of a^{3} is

(A) (1/6)a^{6}

(B) 6a^{4}

(C) (1/6)a^{2}

(D) 6a^{2}

**Solution 2 :**

**Question 1 :**

a, b, c, d are real numbers such that a – 2005 = b + 2006 = c – 2007 = d + 2008. The greatest among a, b, c, d is

(A) a

(B) b

(C) c

(D) d

**Solution 1 :**

**Question 2 :**

a, b, c, d are real numbers such that a – 2005 = b + 2006 = c – 2007 = d + 2008. The greatest among a, b, c, d is

(A) a

(B) b

(C) c

(D) d

**Solution 2 :**

**Question 1 :**

In the adjoining figure, *l*_{1} and *l*_{2} are parallel lines. T is transversal which cuts *l*_{1}, *l*_{2} at A and B respectively. The angles at A, B (refer figure) are trisected. The measure of the angles ACB and ADB are respectively.

(A) 60°, 120°

(B) 120°, 60°

(C) x – y, x + y

(D) 2(x – y), (x + y)

**Solution 1 :**

**Question 2 :**

The number of digits when 200^{8} is written in the decimal form.

(A) 2008

(B) 1004

(C) 74

(D) 19

**Solution 2 :**

**Question 1 :**

The number of three digit numbers whose digits are even is

(A) 64

(B) 75

(C) 100

(D) 125

**Solution 1 :**

**Question 2 :**

ABCD is a square and BCE is an equilateral triangle constructed externally. The measure of ∠AED is

(A) 30°

(B) 20°

(C) 35°

(D) 15°

**Solution 2 :**

**Question 1 :**

Given a sequence of two digit numbers grouped in brackets as follows (10), (11, 20), (12, 21, 30), (13, 23, 31, 40), …(89, 98), (99). The digital sum of the numbers in the brackets having maximum numbers is

(A) 9

(B) 10

(C) 9 or 10

(D) 18

**Solution 1 :**

**Question 2 :**

Using the digits 2 and 7, and addition or subtraction operations only, the number 2010 is written. The maximum number of 7 that can be used, so that the total numbers used is a minimum is

(A) 284

(B) 286

(C) 288

(D) 290

**Solution 2 :**

**Question 1 :**

Given 63.63 = m(21 + n/100), m, n positive integers with n < 100. The value of (m + n) is

(A) 21

(B) 24

(C) 104

(D) 101

**Solution 1 :**

**Question 2 :**

n is a natural number and (n + 2)(n + 4) is odd. Then the biggest power of 2 that divides (n + 1)(n + 3) for any n is

(A) 1

(B) 2

(C) 3

(D) 4

**Solution 2 :**

**Question 1 :**

p is an odd prime. Then is

(A) a fraction less than p

(B) a fraction greater than p

(C) a natural number NOT equal to p

(D) a natural number equal to p

**Solution 1 :**

**Question 2 :**

PQRS is a parallelogram. MP and NP divide ∠SPQ into three equal parts (∠MPQ > ∠NPQ) and MQ and NQ divide ∠RQP into three equal parts (∠MQP > ∠NQP). If k(∠PNQ) = (∠PMQ), then k =

(A) 1/2

(B) 1a fraction greater than p

(C) 3/2

(D) 1/3

**Solution 2 :**

**Question 1 :**

In the adjoining incomplete magic square, the sum of all numbers in any row or column or diagonals is a constant value. The value of x is

(A) 18

(B) 23

(C) 22

(D) 16

**Solution 1 :**

**Question 2 :**

The sum of three different prime numbers is 40. What is the difference between the two biggest ones among them

(A) 8

(B) 12

(C) 20

(D) 24

**Solution 2 :**

**Question 1 :**

If 150% of a certain number is 300, then 30% of the number is

(A) 18

(B) 23

(C) 22

(D) 16

**Solution 1 :**

**Question 2 :**

The sum of five distinct non-negative integers is 90. What can be the second largest number of the five at most?

(A) 82

(B) 43

(C) 34

(D) 73

**Solution 2 :**

**Question 1 :**

The ratio of the angles of a quadrilateral are 7 : 9 : 10 : 10. Then

(A) One angle of the quadrilateral is greater than 120°.

(B) Only one angle of the quadrilateral is 90°.

(C) The sum of some two angles of the quadrilateral is 100°.

(D) There are exactly two right angles as interior angles.

**Solution 1 :**

**Question 2 :**

Three different integers have a sum 1 and product 36. Then

(A) Certainly all of them are positive

(B) Only one is negative

(C) Exactly two of them are negative

(D) All the three are negative

**Solution 2 :**

**Question 1 :**

On the square ABCD , point E lies on the side AD and F lies on BC so that BE = EF = FD = 30 cm. The area of the square (in square cms) is

(A) 300

(B) 900

(C) 800

(D) None of these

**Solution 1 :**

**Question 2 :**

If 2016 = 2^{x}3^{y}5^{z}7^{u} where x, y, z, u are non-negative integers, the value of x + y + 2016z + 3u is

(A) 10

(B) 11

(C) 12

(D) 13

**Solution 2 :**

**Question 1 :**

The fraction 4/37 is written in the decimal form 0.a_{1}a_{2}a_{3}…. The value of a_{2017} is

(A) 8

(B) 0

(C) 1

(D) 5

**Solution 1 :**

**Question 2 :**

The number of integers x satisfying the equation (x^{2} – 3x + 1)^{x + 1} = 1 is

(A) 2

(B) 3

(C) 4

(D) 5

**Solution 2 :**

**Question 1 :**

The fraction greater than 8 4/9 is

(A) 8 1/3

(B) 150/18

(C) 8 2/3

(D) 216/27

**Solution 1 :**

**Question 2 :**

A car is slowly driven in the road full of fog. The car passes a man who was walking at the rate of 3 km an hour in the same direction. He could see the car for 4 minutes and was visible for up to a distance of 100 meters. The speed of the car is (in km per hour)

(A) 8 1/3

(B) 150/18

(C) 8 2/3

(D) 216/27

**Solution 2 :**

**Question 1 :**

If 4921 × D = ABBBD, then the sum of the digits of ABBBD × D is ______.

(A) 19

(B) 20

(C) 25

(D) 26

**Solution 1 :**

**Question 2 :**

What is the 2019^{th} digit to the right of the decimal point, in the decimal representation of 5/28?

(A) 2

(B) 4

(C) 8

(D) 7

**Solution 2 :**

After clearing NMTC Stage 1 Sub Junior Level exam you need to prepare for NMTC Stage 2 Sub Junior level exam. For this, you need **NMTC Stage 2 Sub Junior Level Previous Years Questions Papers with Solutions**.

**Question 1 :**

The greatest common divisor of a and 72 is (a, 72) = 24 and the least common multiples of b and 24 is [b, 24] = 72. Find the GCD (a, b) and LCM [a, b] given that a is the smallest three digit number having this property and b is the biggest integer having this property.

**Solution 1 :**

**Question 1 :**

Let N be the greatest integral multiple of 8 such that no two of its digits are same. What is the remainder when N is divided by 1000?

**Solution 1 :**

**Question 1 :**

If a + 1/a = -1, then find the value of (a^{2} + a + 1) + (a^{4} + a^{2} + 1) + (a^{6} + a^{3} + 1) + (a^{8} + a^{4} + 1) + … + (a^{2006} + a^{1003} + 1).

**Solution 1 :**

**Question 1 :**

The sum of 2000 numbers is 2007. Find the maximum product of these numbers. [There are more than one set of numbers].

**Solution 1 :**

**Question 1 :**

A Predator beast weighs 2008 kg at the beginning of a year. During the first month of the year its weighs is increased by 33 1/3% and in the second month decreased by 25% and in the third month increased by 50%, in the fourth month decreased by 33 1/3%, in the fifth month increased by 16 2/3% and in the sixth month decreased by 21 3/7%. This increase and decrease of the weight continued in the next 6 months in the same order with the same percentage. Find the weight of the predator at the end of the year.

**Solution 1 :**

**Question 1 :**

Find the number of distinct remainders when 2009 is divided by natural numbers. Note that the remainder r is such that 0 ≤ r < 2009.

**Solution 1 :**

**Question 1 :**

a and b are positive integers with a > b. Find all pairs a, b such that the sum of their sum, difference, product and quotient is 36

**Solution 1 :**

**Question 1 :**

Find all the three digit and four digit natural numbers such that the product of the digits is a prime number. Find the sum of all such three digit numbers and the sum of all such four digit numbers. Find the biggest prime factor of each sum.

**Solution 1 :**

**Question 1 :**

Find the number of numbers coprime to and less than 2012. Find their sum. Find also the quotient when this sum is divided by 2012. (Information: 503 is a prime).

**Solution 1 :**

**Question 2 :**

The ratio between two digit number and the sum of the digits of that number is a : b. If the digit in the unit place is n more than the digits in the tens place, prove that the number is given by 9na/(11b – 2a).

**Solution 2 :**

**Question 1 :**

The ratio between two digit number and the sum of the digits of that number is a : b. If the digit in the unit place is n more than the digits in the tens place, prove that the number is given by 9na/(11b – 2a).

**Solution 1 :**

**Question 1 :**

The diagram below contains 13 boxes. The numbers in the second and the twelfth boxes are respectively 175 and 70. Fill up in the boxes with natural numbers such that

- Sum of all numbers in all the 13 boxes is 2015.
- Sum of the numbers in any three consecutive boxes is always the same.

The solution must contain the steps how you arrive at the numbers

**Solution 1 :**

**Question 1 (a):**

If x/a = y/b = z/c = 2016, where x, y, z, a, b, c are non-zero real numbers, find the value of

**Solution 1 (a):**

**Question 1(a) :**

Find all three-digit numbers in which any two adjacent digits differ by 3

**Solution 1(a) :**

**Question 2 :**

In the figure given, ∠A, ∠B and ∠C are right angles. If and ∠AEB = 40° and ∠BED = ∠BDE, then find ∠CDE.

**Solution 2 :**

**Question 4 :**

In a triangle XYZ, the medians drawn through X and Y are perpendicular. Then, show that XY is the smallest side of the triangle XYZ

**Solution 4 :**

To get the **EBook PDF** of **NMTC 2004 to 2019 Stage 1 and 2 for Sub junior Level Questions Papers With Answer Keys & Solutions**, **Click on Image Below**.

PDF File Size : **64 MB** Number of Pages : **342**

The post NMTC Stage 1 and 2 Sub Junior Level Previous Year Papers with Solutions appeared first on .

]]>The post S Chand ICSE Maths Solutions for Class 10 Chapter 4 Linear Inequations Exercise 4 appeared first on .

]]>Hi students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get * S Chand ICSE Maths Solutions for Class 10 Chapter 4 Linear Inequations in One Variable Exercise 4*. This is the fourth chapter ‘

First, we understand some concepts of Linear Inequations in One Variable in this post.

In mathematics, a statement that contains > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to) sign are called an inequality. The inequality defines the comparison between two numbers or other mathematical expressions.

For example:

(i) *a* < *b* means that *a* is less than *b*.

(ii) *a* > *b* means that *a* is greater than *b*.

(iii) *a* ≤ *b* or *a* ⩽ *b* means that *a* is less than or equal to *b. *

(iv) *a* ≥ *b* or *a* ⩾ *b* means that *a* is greater than or equal to *b.*

(v) *a* ≠ *b* means that *a* is not equal to *b.*

A mathematical statement that indicate the value of variables or an algebraic expression which is not equal to other value are called an inequation. These inequations are written using the inequalities signs, > (greater than), ≥ (greater than or equal to), < (less than), ≤ (less than or equal to), ≠ (not equal to).

For example:

(i) x < 5

(ii) y > 10

(iii) (3x + 6) ≤ 5

(iv) (2y – 9) ≥ 15

(v) (3k – 2) ≠ 5

A mathematical statements of any of the forms ax + b > 0, ax + b ≥ 0, ax + b < 0, ax + b ≤ 0 are called as linear inequations in variable x, where a, b are real numbers and a ≠ 0.

For example:

(i) x < 15

(ii) y > 12

(iii) (3x – 8) ≤ 15

(iv) (2y + 9) ≥ 15

(v) (3k + 2) ≠ 5

For an inequation, the set of elements or values from which the values of the variable are taken is called the replacement set or also known as domain of variable.

Now, the solution of an inequation depends upon the replacement set. So, an inequation may have one, many or no solution, depending upon the replacement set.

The solutions of inequation make a set of values of variable, and it is known as solution set of the inequation.

**Every solution set is a subset of replacement set**.

For example: we need to solve x + 2 > 7.

Depending the replacement set, the solutions set will be different.

If replacement set A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then solution of given inequation, means solution is S = {6, 7, 8, 9, 10}.

If replacement set B = {0, 1, 2, 3, 4, 5, 6, 7}, then solution of given inequation, means solution is S = {6, 7}.

If replacement set C = {0, 1, 2, 3, 4}, then solution of given inequation, means solution is S = φ (No Solution).

There are following inequality rules:

(i) *An inequation is NOT changed if the same number is added to both sides of inequation. *

For example:

x – 5 > 12

⇒ x – 5 + 5 > 12 + 5 (**Adding 5 to both sides**)

⇒ x > 17

(ii) *An inequation is NOT changed if the same number is subtracted from both sides of inequation. *

For example:

y + 12 < 15

⇒ y + 12 – 12 < 15 – 12 (**Subtracting 12 from both sides**)

⇒ y < 3

(iii) *An inequation is NOT changed if the same positive number is multiplied to both sides of inequation. *

For example:

x/5 < 2

⇒ x/5 * 5 < 2 * 5 (**Multiplying 5 to both sides**)

⇒ x < 10

(iv) *An inequation is changed if the same negative number is multiplied to both sides of inequation. And, the inequality sign is reversed.*

For example:

-x/3 < 4

⇒ -x/3 * (-3) > 4 * (-3) (**Multiplying -3 to both sides**)

⇒ x > -12

(v) *An inequation is NOT changed if the same positive number divides both sides of inequation. *

For example:

3x < 15

⇒ 3x / 3 < 15 / 3 (**Both sides are divided by 3**)

⇒ x < 5

(vi) *An inequation is changed if the same negative number divides both sides of inequation. And, the inequality sign is reversed.*

For example:

-3x < 21

⇒ -3x / -3 > 21 / -3 (**Both sides are divided by -3**)

⇒ x > -7

(vii) *Any term of an inequation may be taken to other side with its sign changed without affecting the sign of inequality.*

For example:

x + 10 < 51

⇒ x < 51 – 10 (**10 is moved to RHS and it becomes negative without affecting inequality sign**)

⇒ x < 41

On a bargain counter, the shopkeeper puts labels on various goods showing their prices, Rs. P, where P is real numbers. Write a mathematical sentences for each of the following labels

(a) more than Rs. 7.50

(b) not less than Rs. 10

(c) not more than Rs. 22

(d) less than Rs. 11

**S Chand ICSE Maths Solutions: **

Since the price on label is Rs P, then

(a) more than Rs. 7.50 ⇒ P > 7.50

(b) not less than Rs. 10 ⇒ P ≮ 10 or P ≥ 10

(c) not more than Rs. 22 ⇒ P ≯ 22 or P ≤ 10

(d) less than Rs. 11 ⇒ P < 11

You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7

Fill in the blanks

(a) A = {x : x ≥ -3 } = {……..}

(b) B = {x : x ≤ 1 } = {……..}

**S Chand ICSE Maths Solutions:**

You are given the following numbers : -2.6, 5.1, -3, 0.4, 1.2, -3.1, 4.7

It mean, you are given a replacement set as U = {-3.1, -3, -2.6, 0.4, 1.2, 4.7, 5.1}.

(a) Solution of A = {x : x ≥ -3 } = {**-3, -2.6, 0.4, 1.2, 4.7, 5.1**}

(b) Solution of B = {x : x ≤ 1 } = {**-3.1, -3, -2.6, 0.4**}

If the replacement set is {-2, -1, +1, +2, +4, +5, +9}, what is the solution set of each of the following mathematical senetences.

(a) x + 3/2 > 5/2

(b) x – 4 = -3

(c) 2x – 5 ≥ 10

(d) 3y/2 ≤ 5/2

(e) 4x^{2} = 16.

**S Chand ICSE Maths Solutions:**

Since the replacement set is {-2, -1, +1, +2, +4, +5, +9}, then

(a) Solution set of x + 3/2 > 5/2 ⇒ x > 5/2 – 3/2 ⇒ x > 1 is {+2, +4, +5, +9}.

(b) Solution set of x – 4 = -3 ⇒ x = -3 + 4 ⇒ x = 1 is {+1}.

(c) Solution set of 2x – 5 ≥ 10 ⇒ 2x ≥ 15 ⇒ x ≥ 7.5 is {+9}.

(d) Solution set of 3y/2 ≤ 5/2 ⇒ 3y ≤ 5 ⇒ y ≤ 5/3 ⇒ y ≤ 1.66 is {-2, -1, +1}.

(e) Solution set of 4x^{2} = 16 ⇒ x^{2} = 4 ⇒ x = +2 or -2 is {-2, +2}.

List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.

**S Chand ICSE Maths Solutions:**

Since the replacement set is a positive integer, then

Solution set of

30 – 4(2x – 1) < 30

⇒ -4(2x – 1) < 30 – 30

⇒ -4(2x – 1) < 0

⇒ (2x – 1) > 0

⇒ x > 1/2

⇒ x > 0.5

is x = {1, 2, 3, 4, 5,…}

If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, what is the solution set of each of the following mathematical senetences.

(a) x + 4/3 = 7/3

(b) 2x + 1 < 3

(c) x – 6 > 10 – 6

(d) x + 5 = 20

(e) 2x + 3 ≥ 17

**S Chand ICSE Maths Solutions:**

Since the replacement set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then

(a) Solution set of x + 4/3 = 7/3 ⇒ x = 3/3 ⇒ x = 1 is {1}.

(b) Solution set of 2x + 1 < 3 ⇒ 2x < 2 ⇒ x < 1 is {0}.

(c) Solution set of x – 6 > 10 – 6 ⇒ x > 10 is φ.

(d) Solution set of x + 5 = 20 ⇒ x = 15 is φ.

(e) Solution set of 2x + 3 ≥ 17 ⇒ 2x ≥ 14 ⇒ x ≥ 7 is {7, 8, 9}.

(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}. Form all ordered pairs (x, y) such that x is a factor of y and x < y.

(b) Find the truth set of the inequality x > y + 2, where (x, y) ∈ {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)}

**S Chand ICSE Maths Solutions:**

(a) x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}.

(x, y) = x is a factor of y and x < y.

2 is a factor of 4, 6, 18, 54 and 2 < 4, 2 < 6, 2 < 18, 2 < 54, then (x, y) forms the ordered pairs as (2, 4), (2, 6), (2, 18), (2, 54).

4 is a factor of 4 but 4 = 4, then (4, 4) is not in (x, y).

6 is a factor of 18, 54 and 6 < 18, 6 < 54, then (x, y) forms the ordered pairs as (6, 18), (6, 54).

9 is a factor of 18, 27, 54 and 9 < 18, 9 < 27, 9 < 54 then (x, y) forms the ordered pairs as (9, 18), (9, 27), (9, 54).

Thus, the solution set of (x, y) is {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}.

(b) Given inequality is x > y + 2. It means x is greater than y + 2.

In {(1, 2), (2, 3), (5, 1), (7, 3), (5, 6)}, we see that (5, 1) and (7, 3) are two pairs which satisfy the given inequality.

Thus, the solution set of (x, y) is {(5, 1), (7, 3)}.

Find out the truth set of the following open sentences replacement sets are given

(i) 5/x > 7 ; {1, 2}

(ii) 5/x > 2 ; {1, 2, 3, 4, 5, 6}

(iii) x^{2} = 9 ; {-3, -2, -1, 1, 2, 3}

(iv) x + 1/x = 2 ; {0, 1, 2, 3}

(v) 3x^{2} < 2x ; {-4, -3, -2, -1, 0, 1, 2, 3, 4}

(vi) 2(x – 3) < 1 ; {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

**S Chand ICSE Maths Solutions:**

(i) Given replacement set is {1, 2}.

Now given inequality is 5/x > 7 ⇒ x < 5/7 ⇒ x < 0.71

Thus, the solution set of the given inequality is x ∈ φ.

(ii) Given replacement set is {1, 2, 3, 4, 5, 6}.

Now given inequality is 5/x > 2 ⇒ x < 5/2 ⇒ x < 2.5

Thus, the solution set of the given inequality is x ∈ {1, 2}.

(iii) Given replacement set is {-3, -2, -1, 1, 2, 3}.

Now given equation is x^{2} = 9 ⇒ x = 3 or -3.

Thus, the solution set of the given equation is x ∈ {-3, 3}.

(iv) Given replacement set is {0, 1, 2, 3}.

Now given equation is x + 1/x = 2 ⇒ (x^{2} + 1) = 2x ⇒ (x^{2} -2x + 1) = 0 ⇒ (x – 1)^{2} = 0 ⇒ (x – 1) = 0 ⇒ x = 1.

Thus, the solution set of the given equation is x ∈ {1}.

(v) Given replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}.

Now given inequality is 3x^{2} < 2x ⇒ 3x^{2} – 2x < 0 ⇒ x(3x – 2) < 0 ⇒ 0 < x < 2/3 ⇒ 0 < x < 0.66

Thus, the solution set of the given inequality is x ∈ φ.

(vi) Given replacement set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

Now given inequality is 2(x – 3) < 1 ⇒ x – 3 < 0.5 ⇒ x < 3.5.

Thus, the solution set of the given inequality is x ∈ {1, 2, 3}.

**Statement**: The sum of the lengths of any two sides of a triangle is always greater than the length of its third side.

Let x, x+1, x+2 be the lengths of the three sides of a triangle.

(i) Write down the three inequations in x, each of which represents the given statement.

(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.

**S Chand ICSE Maths Solutions:**

The lengths of the three sides of a triangle are x, x+1, x+2.

(i)

Since the sum of the lengths of any two sides of a triangle is always greater than the length of its third side, we have three inequality as below.

x + (x + 1) > (x + 2) ⇒ 2x + 1 > x + 2 ⇒ x > 1 …(1)

x + (x + 2) > (x + 1) ⇒ 2x + 2 > x + 1 ⇒ x > -1 …(2)

(x + 1) + (x + 2) > x ⇒ 2x + 3 > x ⇒ x > -3 …(3)

On combining (1), (2), (3), we get x > 1.

(ii) Since replacement set of x is an integer, means x ∈ {…, -3, -2, -1, 0, 1, 2, 3, …}.

The solution set of x > 1 is x ∈ { 2, 3, 4, …}.

**Answer True or False**

(a) If x + 10 = y + 14, then x > y

(b) |-4| – 4 = 8

(c) If 10 – x > 3, then x < 7

(d) If p = q + 2, then p > q

(e) If a and b are two negative integers such that a < b, then 1/a < 1/b

(f) 3 ∈ {x : 3x – 2 ≥ 5}

**S Chand ICSE Maths Solutions:**

(a) x + 10 = y + 14 ⇒ x – y = 14 – 10 ⇒ x – y = 4 ⇒ x – y > 0 ⇒ x > y.

Thus, the given statement “If x + 10 = y + 14, then x > y” is **TRUE**.

(b) |-4| – 4 = 4 – 4 = 0

Thus, the given statement “|-4| – 4 = 8” is **FALSE**.

(c) 10 – x > 3 ⇒ 10 – 3 > x ⇒ 7 > x ⇒ x > 7

Thus, the given statement “If 10 – x > 3, then x < 7” is **TRUE**.

(d) p = q + 2 ⇒ p – q = 2 ⇒ p – q > 0 ⇒ p > q

Thus, the given statement “If p = q + 2, then p > q” is **TRUE**.

(e) a < b ⇒ 1/a > 1/b

Thus, the given statement “If a and b are two negative integers such that a < b, then 1/a < 1/b” is **FALSE**.

(f) 3x – 2 ≥ 5 ⇒ 3x ≥ 7 ⇒ x ≥ 7/3 ⇒ x ≥ 2.3

Thus, the given statement “3 ∈ {x : 3x – 2 ≥ 5}” is **TRUE**.

Find the solution of the inequation 2 ≤ 2p – 3 ≤ 5. Hence graph the solution set on the number line.

**S Chand ICSE Maths Solutions:**

2 ≤ 2p – 3 ≤ 5 ⇒ 5 ≤ 2p ≤ 8 ⇒ 5/2 ≤ p ≤ 4 ⇒ 2.5 ≤ p ≤ 4

The solution on the number line is as below.

If x is a negative integer, find the solution set of 2/3 + (x + 1)/3 > 0.

**S Chand ICSE Maths Solutions:**

2/3 + (x + 1)/3 > 0 ⇒ (x + 1)/3 > -2/3 ⇒ x + 1 > -2 ⇒ x > -3 …(1)

Since the replacement set of x is negative integer, it means x ∈ {…, -5, -4, -3, -2, -1}.

From (1), x ∈ {-2, -1}.

Write open mathematical sentences, using x for the variable whose graphs would be

**S Chand ICSE Maths Solutions:**

(i)

From the graph, we see that the shaded arrow is left of -2 (included) and all the real numbers are taken up to -2.

Thus, required mathematical sentence is {x : x ≤ -2 and x ∈ R}.

(ii)

From the graph, we see that the shaded arrow is left of 4 (included) and all the real numbers are taken up to 4.

Thus, required mathematical sentence is {x : x ≤ 4 and x ∈ R}.

(iii)

From the graph, we see that the solid circles are at 4 and 5 and only natural numbers are taken.

Thus, required mathematical sentence is {x : 4 ≤ x ≤ 5 and x ∈ N}.

(iv)

From the graph, we see that the solid circles are at 1, 3 and 5 and only natural numbers are taken.

Thus, required mathematical sentence is {x : 1 ≤ x ≤ 5 and x ∈ N}.

(v)

From the graph, we see that the shaded arrow is right of -2 (excluded) and all the real numbers are ahead of -2.

Thus, required mathematical sentence is {x : x > -2 and x ∈ R}.

Answer **TRUE** or **FALSE**:

(i) If 2 – x < 0, then x > 2.

(ii) The graph of the inequations y ≤ 2x includes the origin.

**S Chand ICSE Maths Solutions:**

(i) 2 – x < 0 ⇒ 2 < x.

Thus, the given statement is **TRUE**.

(ii) On putting x = 0 and y = 0 in y ≤ 2x, we get that 0 = 0, which is true.

Thus, the given statement is **TRUE**.

If 25 – 4x ≤ 16, then find

(i) the smallest value of x when x is real number.

(ii) the smallest values of x when x is an integer.

**S Chand ICSE Maths Solutions:**

25 – 4x ≤ 16 ⇒ 25 – 16 ≤ 4x ⇒ 9 ≤ 4x ⇒ x ≥ 9/4.

(i) Thus, the smallest value of x is 9/4 = 2 1/4.

(ii) The smallest integer greater than x ≥ 9/4 is 3.

Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, find the values of x for which -3 < 2x – 1< x + 4

**S Chand ICSE Maths Solutions:**

Given replacement set is x ∈ {1, 2, 3, 4, 5, 6, 7, 9}.

Given inequation is -3 < 2x – 1 < x + 4.

On solving -3 < 2x – 1 ⇒ 2x > -3 + 1 ⇒ 2x > -2 ⇒ x > -1 … (1)

On solving 2x – 1 < x + 4 ⇒ 2x – x < 1 + 4 ⇒ x < 5 …(2)

From (1) and (2), we get -1 < x < 5.

Thus, the solution set of the given inequation is x ∈ {1, 2, 3, 4}.

Solve the inequality 2x – 10 < 3x – 15

**S Chand ICSE Maths Solutions:**

Given inequality is 2x – 10 < 3x – 15

⇒ 2x – 3x < 10 – 15

⇒ -x < -5

⇒ x > 5

Solve the inequation 3 – 2x ≥ x – 12, given that x ∈ N.

**S Chand ICSE Maths Solutions:**

Given inequation is 3 – 2x ≥ x – 12

⇒ -x – 2x ≥ -3 – 12

⇒ -3x ≥ -15

⇒ x ≤ 5

Since the replacement of x is natural number, then the solution set of the given inequation is

x ∈ {1, 2, 3, 4, 5}.

x ∈ {real numbers} and -1 < 3 – 2x ≤ 7, evaluate x and represent it on number line.

**S Chand ICSE Maths Solutions:**

Given inequation is -1 < 3 – 2x ≤ 7

From -1 < 3 – 2x,

⇒ -1 < 3 – 2x ⇒ -1 – 3 < – 2x ⇒ -4 < -2x ⇒ 2x < 4 ⇒ x < 2 …(1)

and 3 – 2x ≤ 7

⇒ 3 – 7 ≤ 2x ⇒ -4 ≤ 2x ⇒ -2 ≤ x …(2)

On combining (1) and (2), we get -2 ≤ x < 2.

The graph of this inequation is

Find the range of values of x which satisfies and graph these values of x on number line.

**S Chand ICSE Maths Solutions:**

Given inequation is

On simplifying, we get

Now the graph of these values of x is as below.

Solution and graph the solution set of

(a)

(b)

**S Chand ICSE Maths Solutions:**

(a) Given inequation is

On solving first inequation, we get 6 ≥ 2 – x ⇒ x ≥ 2 – 6 ⇒ x ≥ -4 … (1)

On solving second inequation, we get x/3 + 2 < 3 ⇒ x/3 < 3 – 2 ⇒ x/3 < 1 ⇒ x < 3 … (2)

On combining (1) and (2), we get -4 ≤ x < 3.

The graph of this inequation is

(b) Given inequation is

On solving first inequation, we get

x/2 < (6 – x)/4 ⇒ 2x < 6 – x ⇒ 3x < 6 ⇒ x < 2… (1)

On solving second inequation, we get

(2 – x)/6 < (7 – x)/9 ⇒ 3(2 – x) < 2(7 – x) ⇒ 6 – 3x < 14 – 2x ⇒ x > -8 … (2)

On combining (1) and (2), we get -8 < x < 2.

The graph of this inequation is

Find the range of values of x which satisfy

Graph these values of x on real number line.

**S Chand ICSE Maths Solutions:**

Given inequation is

On simplifying,we get

Graph of these values of x on number line is as below.

Write down the range of real values of x for which the inequation x > 3 and -2 ≤ x < 5 are both true.

**S Chand ICSE Maths Solutions:**

Graphing the given inequations x > 3 and -2 ≤ x < 5 on same number line, we get

Thus, we get 3 < x < 5.

Solve and graph the solution set of 3x – 4 > 11 or 5 – 2x ≥ 7; x ∈ R.

**S Chand ICSE Maths Solutions:**

On solving the first inequation, we get 3x – 4 > 11 ⇒ 3x > 15 ⇒ x > 5 …(1)

On solving the second inequation, we get 5 – 2x ≥ 7 ⇒ – 2x ≥ 2 ⇒ x ≤ 1 …(2)

On combining (1) and (2), we get x ≤ 1 or x > 5.

The post S Chand ICSE Maths Solutions for Class 10 Chapter 4 Linear Inequations Exercise 4 appeared first on .

]]>The post NCERT Solutions Class 12 Maths Linear Programming Exercise 12.2 appeared first on .

]]>Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Linear Programming Exercise 12.2*.

As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

* NCERT Solutions for class 12* is highly recommended by the experienced teacher for students who are going to appear in

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 1. **

Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost

of the mixture.

**NCERT Solutions**

Let Reshma mix x kg of food P and y kg of food Q. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the mixture must contain atleast 8 units of vitamin A and 11 units of vitamin B.

Thus, we have constraints as 3x + 4y >= 8 and 5x + 2y >= 11.

Now we have total cost of purchasing x kg of food P and y kg of food Q is Z = 60x + 80y.

Then, we need to minimize Z = 60x + 80y by the constraints as 3x + 4y >= 8 and 5x + 2y >= 11.

Now, let the line l_{1} : 3x + 4y = 8 and l_{2} : 5x + 2y = 11 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is unbounded, we need to use Corner Point Method to minimize Z = 60x + 80y.

Now, on solving the equations of lines, we get that the coordinates A(8/3, 0), E(2, 1/2), D(0, 11/2).

Thus, minimum value of Z is 160 at the point (8/3, 0) and (2, 1/2).

Hence, the required minimum cost of mixture is Rs 160.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 2. **

One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

**NCERT Solutions**

Let the number of first kind of cake is x and the number of second kind of cake is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the maximum number of cakes can be made from 5 kg of flour ans 1 kg of fat.

Thus, we have constraints as 200x + 100y <= 5000 or 2x + y <= 50 and 25x + 50y <= 1000 or x + 2y <= 40.

Now we have total number of cakes Z = x + y.

Then, we need to maximize Z = x + y by the constraints as 2x + y <= 50 and x + 2y <= 40.

Now, let the line l_{1} : 2x + y = 50 and l_{2} : x + 2y = 40 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = x + y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(25, 0), E(20, 10) and B(0, 20).

Thus, maximum value of Z is 30 at the point (20, 10).

Hence, the required maximum value of Z = x + y is 30 when 20 cakes of one kind and 10 cakes of another kind is made of ingredients.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 3. **

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

**NCERT Solutions**

Let the number of tennis rackets is x and number of cricket bat is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the availability in the factory is not more than 42 hours of machine time and 24 craftsman time.

Thus, we have constraints as 1.5x + 3y <= 42 or x + 2y <= 28 and 3x + y <= 24.

(i) Total number of bats and rackets made in a factory is Z = x + y.

Then, we need to maximize Z = x + y by the constraints as x + 2y <= 28 and 3x + y <= 24.

Now, let the line l_{1} : x + 2y = 28 and l_{2} : 3x + y = 24 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = x + y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(8, 0), E(4, 12) and B(0, 14).

Thus, maximum value of Z is 16 at the point (4, 12).

Hence, the required maximum value of Z = x + y is 16 when 4 tennis rackets and 12 cricket bats must be made so that the factory works at full capacity.

(ii) Now the profit function is Z = 20x + 10y

Since the maximum value of Z occurred at the point (4, 12),

then Z_{max} = 20(4) + 10(12) = 80 + 120 = Rs. 200 when 4 tennis rackets and 12 cricket bats are made.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 4. **

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his

profit, if he operates his machines for at the most 12 hours a day?

**NCERT Solutions**

Let the number of packets of nuts of cake is x and the number of packets of bolts is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the machine is operated for atmost 12 hours a day.

Thus, we have constraints as x + 3y <= 12 and 3x + y <= 12.

Now, Total profit Z earned on package of nuts and bolts is Z = 17.5x + 7y.

Then, we need to maximize Z = 17.5x + 7y by the constraints as x + 3y <= 12 and 3x + y <= 12.

Now, let the line l_{1} : x + 3y = 12 and l_{2} : 3x + y = 12 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 17.5x + 7y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(4, 0), E(3, 3) and B(0, 4).

Thus, maximum value of Z is 73.5 at the point (3, 3).

Hence, the required maximum value of Z = 17.5x + 7y is 73.5 when the factory produces 3 packages of nuts and 3 packages of bolts.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 5. **

A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

**NCERT Solutions**

Let the number of packages of screws A is x and the number of packages of screws B is y. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the machine is available for the at most 4 hours a day.

Thus, we have constraints as

4x + 6y <= 240 or 2x + 3y <= 120

and 6x + 3y <= 240 or 2x + y <= 80.

Now, Total profit Z earned on packages of screws A and B is Z = 7x + 10y.

Then, we need to maximize Z = 7x + 10y by the constraints as 2x + 3y <= 120 and 2x + y <= 80.

Now, let the line l_{1} : 2x + 3y = 120 and l_{2} : 2x + y = 80 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 7x + 10y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(40, 0), E(30, 20), B(0, 40).

Thus, maximum value of Z is 410 at the point (30, 20).

Hence, the required maximum value of profit Z = 7x + 10y is Rs 410 when the factory produces 30 packages of screws A and 20 packages of screws B.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 6. **

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?.

**NCERT Solutions**

Let the cottage industry manufacturer x pedestal lamp and y wooden shades. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the cutting machine is available for atmost 12 hours and sprayer is available for atmost 20 hours.

Thus, we have constraints as 2x + y <= 12 and 3x + 2y <= 20.

Now, Total profit Z earned is Z = 5x + 3y.

Then, we need to maximize Z = 5x + 3y by the constraints as 2x + y <= 12 and 3x + 2y <= 20.

Now, let the line l_{1} : 2x + y = 12 and l_{2} : 3x + 2y = 20 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 5x + 3y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(3, 0), E(4, 4), C(0, 10).

Thus, maximum value of Z is 32 at the point (4, 4).

Hence, the required maximum value of profit Z = 5x + 3y is Rs 32 when the manufacturer sell 4 lamps and 4 shades.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 7. **

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

**NCERT Solutions**

Let the company manufacture x souvenirs of type A and y souvenirs of type B. Then, x >= 0 and y >= 0.

According to question, we have following data.

It is given that the cutting machine is available for 200 minutes and for assembling is 240 minutes.

Thus, we have constraints as 5x + 8y <=200 and 10x + 8y <= 240 or 5x + 4y <= 120.

Now, Total profit Z earned is Z = 5x + 6y.

Then, we need to maximize Z = 5x + 6y by the constraints as 5x + 8y <=200 and 5x + 4y <= 120.

Now, let the line l_{1} : 5x + 8y = 200 and l_{2} : 5x + 4y = 120 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 5x + 6y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), C(24, 0), E(8, 20), B(0, 25).

Thus, maximum value of Z is 160 at the point (8, 20).

Hence, the required maximum value of profit Z = 5x + 6y is Rs 160 when the company manufacture 8 souvenirs of type A and 20 souvenirs of type B.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.2 Ques No 8. **

A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

**NCERT Solutions**

Let the the merchant stock x desktop computers and y portable computers. Then, x >= 0 and y >= 0.

According to question,

x + y <=250 and 25000x + 40000y <=7000000 or 5x + 8y <= 1400.

Now, Total profit Z earned is Z = 4500x + 5000y.

Then, we need to maximize Z = 4500x + 5000y by the constraints as

x + y <=250, 5x + 8y <= 1400 and x >= 0 and y >= 0.

Now, let the line l_{1} : x + y = 250 and l_{2} : 5x + 8y = 1400 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 4500x + 5000y.

Now, on solving the equations of lines, we get that the coordinates O(0, 0), A(250, 0), E(200, 50), D(0, 175).

Thus, maximum value of Z is 1150000 at the point (200, 50).

Hence, the required maximum value of profit Z = 4500x + 5000y is Rs 1150000 when the company stocks 200 units of desktop model and 50 units of portable model.

The post NCERT Solutions Class 12 Maths Linear Programming Exercise 12.2 appeared first on .

]]>The post NCERT Solutions Class 12 Maths Linear Programming Exercise 12.1 appeared first on .

]]>Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Linear Programming Exercise 12.1*.

As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

* NCERT Solutions for class 12* is highly recommended by the experienced teacher for students who are going to appear in

Solve the following Linear Programming Problems graphically:

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 1.**

Maximise: Z = 3x + 4y

Subject to the Constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.

Now, let the line l : x + y = 4 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is OAB is bounded, we need to use Corner Point Method to maximize Z = 3x + 4y.

Now, on solving the equations x + y = 4, x = 0 and y = 0, we get that the coordinate of O, A and B are (0, 0), (4, 0) and (0, 4) respectively.

Thus, maximum value of Z is 16 at the point (0, 4).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 2.**

Minimise Z = – 3x + 4 y

subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Now, let the line l_{1} : x + 2y = 8 and l_{2} : 3x + 2y = 12 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is OBCE is bounded, we need to use Corner Point Method to minimize Z = – 3x + 4 y.

Now, on solving the equations of lines, we get that the coordinate of O, B, E and C are (0, 0), (0, 4), (2, 3) and (4, 0) respectively.

Thus, minimum value of Z is –12 at the point (4, 0).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 3.**

Maximise Z = 5x + 3y

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

**NCERT Solutions:**

Given Constraints : 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Now, let the line l_{1} : 3x + 5y = 15 and l_{2} : 5x + 2y ≤ 10 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is OBCE is bounded, we need to use Corner Point Method to minimize Z = 5x + 3y.

Now, on solving the equations of lines, we get that the coordinate of O, B, E and C are (0, 0), (0, 3), (20/19, 45/19) and (2, 0) respectively.

Thus, maximum value of Z is 235/19 at the point (20/19, 45/19).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 4.**

Minimise Z = 3x + 5y

subject to : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

Now, let the line l_{1} : x + 3y = 3 and l_{2} : x + y = 2 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is unbounded, we need to use Corner Point Method to minimize Z = 3x + 5y.

Now, on solving the equations of lines, we get that the coordinate of A, E and D are (3, 0), (3/2, 1/2), and (0, 2) respectively.

Thus, minimum value of Z is 7 at the point (3/2, 1/2).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 5.**

Maximise Z = 3x + 2y

subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

Now, let the line l_{1} : x + 2y = 10 and l_{2} : 3x + y = 15 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is bounded, we need to use Corner Point Method to maximize Z = 3x + 2y.

Now, on solving the equations of lines, we get that the coordinate of O, C, E and B are (0, 0), (5, 0), (4, 3) and (0, 5) respectively.

Thus, maximum value of Z is 18 at the point (4, 3).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 6.**

Minimize Z = x + 2y

subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

**NCERT Solutions:**

Given Constraints : 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

Now, let the line l_{1} : 2x + y = 3 and l_{2} : x + 2y = 6 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is unbounded, we need to use Corner Point Method to maximize Z = x + 2y.

Now, on solving the equations of lines, we get that the coordinate of B and C are (0, 3) and (6, 0) respectively.

Thus, minimum value of Z is 6 at the point (6, 0) and (0, 3).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 7.**

Minimise and Maximise Z = 5x + 10 y

Subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

Show that the maximum of Z occurs at two points.

**NCERT Solutions:**

Given Constraints : x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

Now, let the line l_{1} : x + 2y = 120, l_{2} : x + y = 60 and l_{3} : x – 2y = 0 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region CADE is bounded, we need to use Corner Point Method to minimize and maximize Z = 5x + 10y.

Now, on solving the equations of lines, we get that the coordinates as C(60, 0), A(120, 0), D(60, 30), E(40, 20).

Thus, the minimum of Z is 300 at C(60, 0) and maximum of Z is 600 at A(120, 0) and D(60, 30).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 8.**

Minimise and Maximise Z = x + 2y

Subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

Show that the minimum of Z occurs at two points.

**NCERT Solutions:**

Given Constraints : x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

Now, let the line l_{1} : x + 2y = 100, l_{2} : 2x – y = 0 and l_{3} : 2x + y = 200 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region BDCE is bounded, we need to use Corner Point Method to minimize and maximize Z = x + 2y.

Now, on solving the equations of lines, we get that the coordinates as B(0, 50), D(0, 200), C(50, 100), E(20, 40).

Thus, the maximum of Z is 400 at D(0, 200) and minimum of Z is 100 at B(0, 50) and E(20, 40).

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 9.**

Maximize Z = – x + 2y,

Subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

Now, let the line l_{1} : x = 3, l_{2} : x + y = 5 and l_{3} : x + 2y = 6 and draw its graph and also apply the given constraints. The shaded region in the given figure is the feasible region determined by the system of given constraints.

Since the feasible region is unbounded, we need to use Corner Point Method to maximize Z = – x + 2y.

Now, on solving the equations of lines, we get that the coordinates as C(6, 0), E(4, 1), and F(3, 2).

From the table, we see that the maximum value of Z is 1 at F(3, 2). But the feasible region is unbounded, therefore, we draw the graph of the inequality – x + 2y > 1. Since half plane represented by – x + 2y > 1 has points common with the feasible region.

Thus, the maximum value of Z cannot be 1. hence, Z has no maximum value.

**NCERT Solutions for Class 12 Maths Linear Programming ****Exercise**** 12.1 Ques No 10.**

Maximise Z = x + y,

Subject to the constraints: x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

**NCERT Solutions:**

Given Constraints : x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

Now, let the line l_{1} : x – y = –1, and l_{3} : –x + y = 0 and draw its graph and also apply the given constraints.

Since there is no common region, hence there is no feasible region.

Therefore, there is no maximum value of Z.

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]]>Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this article, you get the full details of * Common Core Algebra 1 Unit 4 Functions Chapters List and Summary*. It means, I provide you short details of chapters of

Now, there are following chapters in **Common Core Algebra 1 Unit 4A Functions Concepts**

In this chapter, you will learn about how to graph a relationship.

Graphs can be used to illustrate many different situations.

Some graphs are connected lines or curves called **continuous graphs**.

Some graphs are only distinct points. These are called **discrete graphs**. In this case, the values between two dots are not included as they have no meaning in the given situation.

Lets understand it by solving some questions:

**Common Core Algebra 1 Unit 4A Chapter 1 Question : 1**

Fill in the blank: A ______ graph is made of connected lines or curves. (continuous, discrete)

**Common Core Algebra 1 Unit 4A Chapter 1 Solution : 1**

A ** connected** graph is made of connected lines or curves.

**Common Core Algebra 1 Unit 4A Chapter 1 Question : 2**

Fill in the blank: A ______ graph is made of distinct points. (continuous, discrete)

**Common Core Algebra 1 Unit 4A Chapter 1 Solution : 2**

A ** discrete** graph is made of distinct points.

In this chapter, you will learn about what the relation and function are and how to find the domain and range of the relation and functions.

In previous chapter, you see that a relationship is represented by a graph. The relationships can also be represented by a set of order pairs, called a relation.

A **relation** can also be represented by tables, graphs or mapping diagrams.

Example: The relation {(1, 5), (2, 3), (3, 2), (4, 1)} can be represented as a table, as a graph and as a mapping diagram.

The **domain** of a relation is the set of first coordinate (x-coordinate) of the ordered pairs and the range of a relation is the set of second coordinate (y-coordinate) of the order pairs.

For example: In the relation {(1, 5), (2, 3), (3, 2), (4, 1)}, the domain is {1, 2, 3, 4} and the range is {5, 3, 2, 1}.

A function is a special type of relation that pairs each domain value with exactly one range value.

Lets understand it by solving some questions:

**Common Core Algebra 1 Unit 4A Chapter 2 Question : 1**

What is the domain and range of {(-5, 7), (0, 0), (2, -8), (5, -20)}

**Common Core Algebra 1 Unit 4A Chapter 2 Solution : 1**

Domain of given relation is {-5, 0, 2, 5} and Range of given relation is {7, 0, -8, -20}.

**Common Core Algebra 1 Unit 4A Chapter 2 Question : 2**

What is the domain and range of {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)}.

**Common Core Algebra 1 Unit 4A Chapter 2 Solution : 2**

Domain of given relation is {1, 2, 3, 4, 5} and Range of given relation is {2, 4, 6, 8, 10}.

**Common Core Algebra 1 Unit 4A Chapter 2 Question : 3**

What is the domain and range of the relation as given in table form below.

**Common Core Algebra 1 Unit 4A Chapter 2 Solution : 3**

Domain of given relation is {3, 5, 2, 8, 6} and Range of given relation is {9, 25, 4, 81, 36}.

In this chapter, you will learn about how to identify independent and dependent variables and how to write an equation in function notation and evaluate a function for given input values.

The input of a function is the independent variable. The output of a function is the dependent variable. The value of the dependent variable depends on the value of independent variables.

Hence, the dependent variable is a function of independent variables. If ‘x’ is the independent variable and ‘y’ is the dependent variable, then the function is written as y = f(x).

Lets understand it by solving some questions:

**Common Core Algebra 1 Unit 4A Chapter 3 Question : 1**

For the function f(x) = 5x, find value of f(x) at x = 6 and at x = 7.5.

**Common Core Algebra 1 Unit 4A Chapter 3 Solution : 1**

Given function is f(x) = 5x.

Put x = 6, then f(6) = 5(6) = 30.

Put x = 7.5, then f(7.5) = 5(7.5) = 37.5.

**Common Core Algebra 1 Unit 4A Chapter 3 Question : 2**

For the function f(x) = 2x – 1, find value of f(x) at x = 1 and at x = -3.

**Common Core Algebra 1 Unit 4A Chapter 3 Solution : 2**

Given function is f(x) = 2x – 1.

Put x = 1, then f(1) = 2(1) – 1 = 2 – 1 = 1

Put x = -3, then f(-3) = 2(-3) – 1 = -6 – 1 = -7

**Common Core Algebra 1 Unit 4A Chapter 3 Question : 3**

For the function f(x) = x/4 + 1, find value of f(x) at x = -24 and at x = 400.

**Common Core Algebra 1 Unit 4A Chapter 3 Solution : 3**

Given function is f(x) = x/4 + 1.

Put x = -24, then f(-24) = -24/4 + 1 = -6 + 1 = -5

Put x = 400, then f(400) = 400/4 + 1 = 100 + 1 = 101

In this chapter, you will learn about how to graph the functions for a given limited domain and for all real numbers.

To graph a function for all real numbers, you need to do the following steps:

Step 1: Use the function to generate ordered pairs by choosing several values for x.

Step 2: Plot enough points to see a pattern for the graph.

Step 3: Connect the points with a line or smooth curve.

Lets understand it by solving some questions:

**Common Core Algebra 1 Unit 4A Chapter 4 Question : 1**

Graph the function f(x) = |x| for the domain {-2, -1, 0, 1, 2}.

**Common Core Algebra 1 Unit 4A Chapter 4 Solution : 1**

First, you need to find the values of f(x) at x = -2, -1, 0, 1, 2.

Now, graph the ordered pair.

**Common Core Algebra 1 Unit 4A Chapter 4 Question : 2**

Graph the function y = 2x + 1

**Common Core Algebra 1 Unit 4A Chapter 4 Solution : 2**

First, you need to choose some values of x and find corresponding values of f(x) to generate ordered pairs.

Now, plot the points to see a pattern and then connect all the points by a line.

Now, there are following chapters in **Common Core Algebra 1 Unit 4B Applying Functions**

In this chapter, you will learn about how to create and interpret scatter plots and how to use trend lines to make prediction.

A scatter plot is a graph with points plotted to show a possible relationship between two sets of data. A scatter plot is an efficient way to display some types of data.

A correlation describes a relationship between two data sets. A graph may show the correlation between data. The correlation can help you analyze trends and make the prediction.

There are three types of correlation between the data:

Lets understand it by solving some questions:

**Common Core Algebra 1 Unit 4B Chapter 5 Question : 1**

The table shows the number of species added to the list of endangered and threatened species in the UNITED STATES during the given years. Graph a scatter plot using the given data.

**Common Core Algebra 1 Unit 4B Chapter 5 Solution : 1**

Required scatter plot of the given table data is below.

In this chapter, you will learn about arithmetic sequence and how to find its nth term.

A sequence is a list of numbers that may form a pattern. Each number in a sequence is term.

When the terms of a sequence differ by the same nonzero number * d*, the sequence is an

To find a term in an sequence, you need to add d to its previous term. The formula to find nth term of arithmetic sequence is

ets understand it by solving some questions:

**Common Core Algebra 1 Unit 4B Chapter 6 Question : 1**

Determine whether the sequence appears to be an arithmetic sequence. If so, find the common difference: 12, 8, 4, 0, ….

**Common Core Algebra 1 Unit 4B Chapter 6 Solution : 1**

Given sequence is 12, 8, 4, 0, ….

Since

8 – 12 = -4

4 – 8 = -4

0 – 4 = -4… it means this is in arithmetic sequence and its common difference is -4

**Common Core Algebra 1 Unit 4B Chapter 6 Question : 2**

Find 60th term of the arithmetic sequence : 11, 5, -1, -7, …

**Common Core Algebra 1 Unit 4B Chapter 6 Solution : 2**

Given arithmetic sequence : 11, 5, -1, -7, …

We have a_{1} = 11, d = 5 – 11 = -6 and n = 60.

Using formula a_{n} = a_{1} + (n – 1)d, we get

a_{60} = 11 + (60 – 1)(-6) = 11 – 6 x 59 = – 343

**Common Core Algebra 1 Unit 4B Chapter 6 Question : 3**

Find the common difference for the arithmetic sequence 107, 105, 103, 101, …

**Common Core Algebra 1 Unit 4B Chapter 6 Solution : 3**

Given arithmetic sequence : 107, 105, 103, 101, …

Required difference is d = 105 – 107 = -2

**Common Core Algebra 1 Unit 4B Chapter 6 Question : 4**

Find 17th term of the arithmetic sequence : 5, 10, 15, 20, 25, …

**Common Core Algebra 1 Unit 4B Chapter 6 Solution : 4**

Given arithmetic sequence : 5, 10, 15, 20, 25, …

We have a_{1} = 5, d = 10 – 5 = 5 and n = 17.

Using formula a_{n} = a_{1} + (n – 1)d, we get

a_{17} = 5 + (17 – 1)(5) = 5 + 80 = 85

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