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 5 of 9.

Jaisri said:   6 years ago
6, 9, 15, 18 LCM is 90.

Least number from this is 6. So 90+6= 96 when 96÷7 is 73 then remainder is 4.

Shalini said:   6 years ago
LCM of 6, 9, 15, 18 is 90.

Such that there is one formula these type of questions ie "LCM*k+remainder "should be divisible by a multiple.

So we want to do "trail and error" method by giving value from the options to the 'k' so that the number is divisible, so we substitute k=4.

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

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

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

Please clear it.

Meenakshi said:   1 decade ago
@bhavi,

The trick which was said by krishna was very simple and even though you ask for a short trick ?

Sowmya said:   1 decade ago
94 is divisible by 7 and it leaves a reminder of 4 when divided with all other given numbers so y not 94?

Koti reddy said:   1 decade ago
How to come k=4 ?

Sachin said:   1 decade ago
Given here the answer should be multiple of 7, so simply divide all options by 7.

Then we only got the answer 364.

Anurag nist said:   1 decade ago
No need to find ok...

Just substitute k=1,k=2,k=3,....so on.

The question asks for least number....

The moment you get that the num is divisible by 7. stop there....

I hope u got it.


Post your comments here:

Your comments will be displayed after verification.