Find the greatest number which divides 6168 2447 and 3118 leaving the same remainder in each case?

The greatest such number is 1.

If n is such a number, then

6168 = an + r

2447 = bn + r, and

3118 = cn + r for some integers a, b, c and r.

This means that

6168 - 2447 = 3721 = (a-b)n

6168 - 3118 = 3050 = (a-c)n, and

3118 - 2447 = 671 = (c-b)n

That is, n is the greatest common factor of 3721, 3050 and 671.

But the GCF of these numbers is 1. Hence the answer.