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
Answer: Option
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:
88 comments Page 8 of 9.

Aqsa said:   1 decade ago
@ Krishan Rajwar u made me laugh.u r good.

Adi said:   7 years ago
How 90k+4 ?

Please explain this step.

Choudhary said:   6 years ago
What if the value of k has very large?
(1)

Neetu said:   1 decade ago
90k+4/7=4

How is please explain it.

Priya said:   9 years ago
How Lcm of 90 come? Please explain.

Sherin lazar said:   4 months ago
Thank you all for explaining answer.

Dashi said:   7 years ago
Thanks for the answer @Jignesh.

Codi said:   3 years ago
Thanks for explaining @Rahul.
(2)

SATYA said:   9 years ago
How find the value of k==4?

Jyoti said:   1 decade ago
How the value k=4 comes?


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