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

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

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

Answer»

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

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