Question

find the smallest number which when divided by 28 and 32 leaves remainder 3 and 12 respectively

Answer 1

given that the smallest number divided by 28 and 32 leaves the remainder 3 and 12.

28 - 8 = 20 and 32 - 12 = 20 divided by the required number.


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 - 198

= 8.

204/32 = 32 * 6 = 192 = 204 - 192

= 12.

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