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 7 of 9.

Sundar said:   1 decade ago
Four different numbers are given. So take k=4 that's all.

Sherin lazar said:   4 months ago
Thank you everyone for explaining the session clearly.

Jambay said:   5 years ago
Why we are taking LCM instead of GCF? Please explain.
(7)

Jpbernela said:   1 decade ago
How can you find the value of k=4 ? conclude please.

Manisha said:   1 decade ago
Please, tell the how you calculate the value of k=4?

Gurpreet singh said:   9 years ago
I understand you trick, Thank you @Jignesh Rajput.

Neetu said:   1 decade ago
How k's value is 4 9k+4/7= k=4.

Please clear it.

Nakum pragnesh said:   3 years ago
Or which option is divisible by 7 is your answer
(23)

Hira said:   10 years ago
Correct answer is 94. Read question carefully.

Mohsin said:   1 decade ago
How did you get 90? Can anyone please explain.


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