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Oct 31, 2019 in Other
Q:

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

1 Answer

Oct 31, 2019
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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