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