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

Here is my simple explanation.

I think everyone is understanding how did they get ab = 12,
now if first no. is 13a,
and second no. is 13b,
factors of 12 can be (1,12) (3,4) (6,2),

Now, see if you include any common factor in a and b both then that will increase the H.C.F as HCF if highest common factor like if we take factor 6 and 2 then there is common factor between 6 and 2 which is 2 so HCF will change now to 26 that's why we can't take 6 and 2.
So only 2 pairs (1,12) (3,4).

Answer = 7.
Remainder = 4,
7*4 =28.
which answer divide by 28 that is the correct answer is it right?

How come 90k+ 4?

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.

How 90k+4 ?

Please explain this step.

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.

Don't go for calculation first here it is told that multiple of 7. Here is your answer.

I didn't understand this problem.

Please anyone explain this problem clearly.

It should be 94 which is divisible by 7 as well as leaves reminder 4 for other number.

Thanks all.