Question

Find the smallest no which when divided by 28 and 32 leaves remanider 8 and 12 respictively

Answer 1

Hey !

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 * 7

Prime factorization of 32 = 2 * 2 * 2 * 2 * 2

LCM(28,32) = 2 * 2 * 2 * 2 * 2 * 7

                     = 224.

Therefore the required smallest number = 224 - 20

                                                                    = 204.

Verification is in the attachment

Hope it Helps