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

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

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.

Here is a simple logic.
First find the LCM of the no.s which comes 90.
Now find the no. among the options greater than 90 and divisible by 7.

Isn't it the least common multiple of 6, 7, 9, 15 and 18 with a remainder 4?

You will have 90k+4.

Now we know 90/7 gives - 12.85 so, we can write 90 as 84k+6k.

So 90k+4 can be written as (84k)+6k+4. The first part 84k is divisible by 7. So no need to check. The second part i.e. 6k+4 has to be divisible by 7.

K = 1, 6(1)+4 = 10 X.

K = 2, 6(2)+4 = 16 X.

K = 3, 6(3)+4 = 22 X.

K = 4, 6(4)+4 = 28.

So this is divisible by 7 hence your answer is 84(4)+6(4)+4.

Hope this clarifies.

Shortest Way => Only one option divisible by 7 is "364".

k=4 because we need least value of k which is divided 7.

So in equation 90k+4. If k=4 then answer is 364 which is divisible by 7.

Please, tell the how you calculate the value of k=4?

Four different numbers are given. So take k=4 that's all.

It is given in the question that it should be least multiple of 7 which leaves remainder 4. So why don't it be 94. Please clarify.