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

I didn't understand this problem.

Please anyone explain this problem clearly.

Thanks for the answer @Jignesh.

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.

How 90k+4 ?

Please explain this step.

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 come 90k+ 4?

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

Here you can check from option also which is divisible of 7 like;
74 is not completely divided by 7.

Same as 94 and 184 also, But 364 is completely divisible by 7, So 364 is the answer.

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).

By formula;.

Given reminder is 4, and divisor is 6, 9, 15, 18 and lcm of 6, 9, 15, 18 is 90. To find dividend we have to multiply 90 with remainder 4 and then add remainder ie 4.