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:
4
5
6
8
Explanation:
N = H.C.F. of $\left(4665 - 1305\right)$, $\left(6905 - 4665\right)$ and $\left(6905 - 1305\right)$
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = $\left( 1 + 1 + 2 + 0 \right)$ = 4