find the smallest no which when divided by 28 and 32 leaves remainder 8 and 12 respectively Question find the smallest no which when divided by 28 and 32 leaves remainder 8 and 12 respectively in progress 0 Math Savannah 2 weeks 2021-09-13T22:35:08+00:00 2021-09-13T22:35:08+00:00 1 Answer 0 views 0

## Answers ( )

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 * 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:

204/28 = 28 * 7 = 196.

= 204 – 196

= 8

204/32 = 32 * 6 = 192

= 204 – 192

= 12.

Hope this helps!