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1. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:





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  • By: guest on 01 Jun 2017 05.48 pm
    L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is multiple of 7. Least value of k for which (90k + 4) is divisible by 7 is k = 4. Required number = (90 x 4) + 4   = 364.
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