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

let the greatest number be a and remainder be x. And let the whole number result of the division be b, c and d. So we have

ab + x = 2447 ... (1)

ac + x = 3118 ... (2)

ad + x = 6168 ... (3)

Subtracting (2) - (1) gives:-

ac - ab = 671

a (c - b) = 671

Now 671 = 61 * 11 so 61 could be the value of a

Checking: - 2447 / 61 = 40 remainder 7, 3118 / 61 = 51 remainder 7

and 6168 / 61 = 101 remainder 7

Answer 61