Aptitude - Problems on H.C.F and L.C.M - Discussion

So basically this is kind of a trial an error method question you just need to find the value of k such that 90k+4/7 gives 0.

But in this question, the least number is 4. However, what if the number is too large. Will we have to try with every number in place of k and keep checking if it is divisible on not?

L.C.M. of 6, 9, 15 and 18 is 90.

Take k =1,2,3,... till reminder is 00.

Let required number be (90 k + 4 )/7.

if k=1 then 90(1)+4/7 = 94/7 =13.42 = reminder is not 00.

if k=2 then 90(2)+4/7 = 184/7 =26.28 = reminder is not 00.

if k=3 then 90(3)+4/7 = 274/7 =39.14 = reminder is not 00.

if k=4 then 90(4)+4/7 = 364/7 =52.00 = reminder is 00.

OK now you take k = 4 because remainder 0.
Solution is 90k + 4 = 90*4 + 4 = 364.


Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

a/b ve get quocent q and reminder r
there for a=bq+r that why 90k+4

Let required number be 90k + 4, which is multiple of 7.

How we are taking 90k+4??

Thank rahul.

@bhavi,

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

94 is divisible by 7 and it leaves a reminder of 4 when divided with all other given numbers so y not 94?

How to come k=4 ?

Given here the answer should be multiple of 7, so simply divide all options by 7.

Then we only got the answer 364.