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 6 of 9.

Sakthivel C.M said:   1 decade ago
How did you find K ?

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

Manasa said:   1 decade ago
Thank you sundar.

Sundar said:   1 decade ago
Short cut method: [ Eliminating the options method ]

From this "The least multiple of 7" - we can tell the the result should be divisible by 7.

From the given 4 options, 364 is the only number divisible by 7.

Therefore, 364 is the correct answer.

Jignesh rajput said:   1 decade ago
Here, (90k+4) put each of value starting from 1,2,3,.... than after divisible by 7.when in the case of put 4th value we have not get the any remainder of (90k+4)divisible by 7..dats y take k=4.

i.e (90(1)+4)/7= remainder 3
(90(2)+4)/7= remainder 3
(90(3)+4)/7= remainder 2
(90(4)+4)/7= remainder 0
thats y take k=4.
thanks u

Akhil said:   1 decade ago
Thank rahul.

Neetu said:   1 decade ago
90k+4/7=4

How is please explain it.

Habibali@mail.com said:   1 decade ago
Thankx Rahul...

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

Bhavi said:   1 decade ago
Is there any short trick to solve these types of questions???


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