The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:	Answers
A) 74 B) 94 C) 184 D) 364 E) 512
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.   Answer: Option 364
<result>D</result>
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: