# Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 10)
10.
The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
74
94
184
364
Explanation:

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.

Discussion:
86 comments Page 9 of 9.

The question is a bit confusing but once you get it, it is ridiculously easy to solve. We are basically looking for a number X that has the following properties:
X/7 - no remained
x/6 - remainder of 4
X/9 - remainder of 4
X/15 - remainder of 4
X/18 - remainder of 4

Ideally, one would try and find the LCM of 6, 9, 15 and 18 but we already know the number involved is a multiple of 7 so we can quickly eliminate by dividing everything by 7. Turns out only one option remains. Check if they all divide by 6, 9, 15 and 18 to give a remainder of 4 and this is one of the easiest questions that can be asked.

How can you find the value of k=4 ? conclude please.

It can not be 1, 2, 3 because if we substitute these numbers the result which is not divisible by 7. Let's check out. The only least number is 4.