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

The smallest number when divided by 28 and 32 leaves remainder 8 and 12 respectively:-

28 – 8 = 20 and 32 – 12 = 20 are divisible by the required numbers.

Therefore, the required number will be 20 less than the LCM of 28 and 32.

Prime factorization of 28 = 2 x 2 x 7 Prime factorization of 32 = 2 x 2 x 2 x 2 x 2

LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224.

Therefore, the required the smallest number = 224 – 20 = 204.

Verification: 204/28 = 28 x 7 = 196.

= 204 – 196 = 8 204/32

= 32 x 6 = 192

= 204 – 192 = 12.

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