Discussion Forum

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case Then sum of the digits in N is :

Answer:

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Note:

We are taking hcf of (4665 - 1305), (6905 - 4665) and (6905 - 1305) because to cancel the remainder.

take an example..
12 and 7, both give us remainder 2 when divided by 5.
now when we substact (12-7) the differance 2 is cancelled.