# What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case?

Given: 35, 56 and 91.

To find: Here we have to find the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case.

Solution:

LCM of two numbers is the smallest number which is exactly divisible by both numbers.

First, we need to find the LCM of 35, 56 and 91.

Now, calculating the LCM of 35, 56 and 91 by the prime factorization method:

Writing the numbers as a product of their prime factors:

Prime factorisation of 35:

• $5\ \times\ 7\ =\ 5^1\ \times\ 7^1$

Prime factorisation of 56:

• $2\ \times\ 2\ \times\ 2\ \times\ 7\ =\ 2^3\ \times\ 7^1$

Prime factorisation of 91:

• $7\ \times\ 13\ =\ 7^1\ \times\ 13^1$

Multiplying the highest power of each prime number:

• $2^3\ \times\ 5^1\ \times\ 7^1\ \times\ 13^1\ =\ 3640$

LCM(35, 56, 91)  $=$  3640

But we have to find the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case. So,

The required number  $=$  LCM(35, 56, 91)  $+$  7

The required number  $=$  3640  $+$  7

The required number  $=$  3647

So, the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case is 3647.

Updated on: 10-Oct-2022

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