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:

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Solution

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 the digits in N = (1 + 1 + 2 + 0) = 4

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Please answer this question

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

N

1305

4665

6905

N

1305, 4665 a n d 6905

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