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# 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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