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

First read the q. when we divide the all options with 6 then we get the reminder 4 in three option's 1st is 94 and 2nd 184 and 3rd is 364. when we divide the 94 and 184 with 7 we will get it is not divisible by 7, but when we divide the 364 with 7 then we get this is divisible by 7 so it will be answer.

6)74(12 6)94(15  6)364(6  6)184(3
 -6      -6       -36      -18
-----   -----    ------   ------
  14      34       004      004
 -12     -30
-----   -----
  02      04

After this only 364 is divisible by 7

7)364(52
-35
------
014
- 14
-------
00
-------

Another way is,
Divide all 4 option with 7. you will easily get which no. is divisible by 7.
And you will get only 364 is divisible by 7 so answer is 364.

I'm confused to find LCM and HCF. Can anyone please explain?

@Mohsin.

Factors of 6 = 2*3.
Factors of 9 = 3*3.
Factors of 15= 3*3*5.
Factors of 18 = 3*2*3.
LCM = 3*2*3*5= 90.

How did you get 90? Can anyone please explain.

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??