The smallest number that leaves 3 as remainder when divided by 28 and 42 is
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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!
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