Question

The smallest number which gives remainders 8 and 12 when divided by 28 and 32 respectively is..

a) 180 b) 240 c) 204 d) 210​

Answer 1

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. LCM(28, 32) = 2 x 2 x 2 x 2 x 2 x 7 = 224. Therefore, the required the smallest number = 224 – 20 = 204.

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