Question

Let Me be the greatest number that will divide 1305,4665 & 6905, leaving the same remainder ineach case. Sum of digits of N is ---------

Answer 1

Sum of digit of N is 4

Let us say that the same remainder in each case is 'r'
1305 - r is divisble by N
4665 - r is divisible by N
6905 - r is divisible by N

If 'a' is divisible by N and 'b' is divisible by N, then 'b - a' is divisible by N
Case 1: (4665 - r) - (1305 - r) is divisble by N
=> 3360 is divisble by N
Case 2: (6905 - r) - (1305 - r) is divisble by N
=> 5600 is divisble by N
Case 3: (6905 - r) - (4665- r) is divisble by N
=> 2240 is divisble by N

We are looking for the largest possible value of N.
It will be the HCF(3360, 5600, 2240) = 1120

Sum of digit= 1+1+2+0=4