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:

Question: 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:

[A].

4

[B].

5

[C].

6

[D].

8

Answer: Option A

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