# 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:
85 comments Page 1 of 9.

Nakum pragnesh said:   9 months ago
(2)

Codi said:   11 months ago
Thanks for explaining @Rahul.

Sagar Khare said:   1 year ago
@ALL.

Guys here you use divisibility rules.

We have to find 7 multiple so apply the simple divisibility rule of 7 on all the answers.
(3)

Bhavesh said:   2 years ago
90K+4 is the required number.

If K=1,
90*1+4=94 , it is not multiple of 7.
90*2+4=184,it is not multiple of 7,
90*3+4=274,it is not multiple of 7,
90*4+4=364,it is multiple of 7,
So, the least value will be 364.
(9)

Navnath said:   2 years ago
As sample as that.

If the number is multiple of 7 then Obviously it should divide by 7.
Then check all the options by dividing 7.

Look only one option that is 364 is completely divided by 7.
(1)

SUBHANKAR DAS said:   2 years ago
L.C.M. of 6, 9, 15, and 18 is 90.
Let the required number be 90k + 4, which is a multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4,as it is not divisible if k=1,2,3
So, the Required number = (90 x 4) + 4 = 364.

Akash said:   2 years ago
@Jeni.

Can you please explain how came 7 when calculating the LCM?
(2)

Jeni said:   3 years ago
6 = 2,3.
9 = 3,3.
15 = 3,5.
18 = 2,3,3.
LCM = 7 x 2 x 3 x 3 x 5 = 360.
Add the reminder = 360 + 4 = 364.
(5)

Jambay said:   3 years ago