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

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