IndiaBIX

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)
Problems on H.C.F and L.C.M
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.
Shakir said:   7 years ago
Here you can check from option also which is divisible of 7 like;
74 is not completely divided by 7.

Same as 94 and 184 also, But 364 is completely divisible by 7, So 364 is the answer.
(1)
Mansi said:   7 years ago
Answer = 7.
Remainder = 4,
7*4 =28.
which answer divide by 28 that is the correct answer is it right?
Sweet said:   7 years ago
How come 90k+ 4?
Denusan said:   7 years ago
When the leaves of the remainder is not equal to get the value of 4 then 90+4=94.

Is it right thew given is used in it.
Adi said:   7 years ago
How 90k+4 ?

Please explain this step.
Sai kowsalya.D said:   7 years ago
K=4 because 4 is d Perfect no that suits to d equation i.e (90k+4).

If we substitute k=4 then it becomes 364 hence 364 is divisible by 7 i.e 52 hence k=4.
Dashi said:   7 years ago
Thanks for the answer @Jignesh.
Smile said:   7 years ago
I didn't understand this problem.

Please anyone explain this problem clearly.
Deb said:   6 years ago
It should be 94 which is divisible by 7 as well as leaves reminder 4 for other number.
Deepak said:   6 years ago
Thanks all.
Post your comments here:

Your comments will be displayed after verification.

Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers