question_answer

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

Correct Answer: A

Solution :

The greatest N = HCF of \[(1350-x),\] \[(4665-x)\] and \[(6905-x),\] where x is the remainder \[=HCF\,\,\text{of}\,\,(4665-1305),\] \[(6905-4665)\] and \[(6905-1305)\] = HCF of 3360, 2240 and 5600             Again,             \[\therefore \]      \[N=1120\] Sum of digits \[=1+1+2+0=4\]