Question

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

• A. 4
• B. 5
• C. 8
• D. 9

Answer 1

The correct option is A

4

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.