# Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Given: 28 and 32.

To find: Here we have to find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Solution:

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

First, we need to find the LCM of 28 and 32.

Now, calculating the LCM of 28 and 32 by the prime factorization method:

Writing the numbers as a product of their prime factors:

Prime factorisation of 28:

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

Prime factorisation of 32:

• $2\ \times\ 2\ \times\ 2\ \times\ 2\ \times\ 2\ =\ 2^5$

Multiplying the highest power of each prime number:

• $2^5\ \times\ 7^1\ =\ 224$

LCM(28, 32)  $=$  224

But we have to find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. So,

The required number  $=$  LCM(28, 32)  $-$  8  $-$  12

The required number  $=$  224  $-$  20

The required number  $=$  204

So, the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively is 204.

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Updated on: 10-Oct-2022

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