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

Given that 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 $28=2\times 2\times 7$

Prime factorization of $32=2\times 2\times 2\times 2\times 2$

$LCM(28,32)=2\times 2\times 2\times 2\times 2\times 7=224$

Therefore the required smallest number $=224-20=204.$