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

Find the smallest number which which divided by 28 and 32 leaves remai

1.	Find the smallest number which which divided by 28 and 32 leaves remainders 8 and 12 respectively.
Answer» 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 × 2 × 7Prime factorization of 32 = 2 × 2 × 2 × 2 × 2LCM(28,32) = 2 × 2 × 2 × 2 × 2 × 7 = 224.Therefore the required smallest number = 224 - 20 = 204.Verification:204/28 = 28 × 7 = 196. = 204 - 196 \xa0 = 8204/32 = 32 × 6 = 192 = 204 - 192 = 12. Answer is 20 204..