Question

N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digit s in n?N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the

Answer 1

The problem asks for the largest N such that:

4665=aN+x4665=aN+x
1305=bN+x1305=bN+x
6905=cN+x6905=cN+x

where x>0x>0.

We can remove xx by subtracting the corresponding sides of any two of these equations:

3360=4665−1305=(a−b)N3360=4665−1305=(a−b)N
5600=6905−1305=(c−b)N5600=6905−1305=(c−b)N
2240=6905−4665=(c−a)N2240=6905−4665=(c−a)N

So, N must be a factor of all three of these new numbers.  Let's pick the largest such factor and make sure it works.  N=gcd(2240,3360,5600)=1120N=gcd(2240,3360,5600)=1120 gives us

4665=4∗1120+1854665=4∗1120+185
1305=1∗1120+1851305=1∗1120+185
6905=6∗1120+1856905=6∗1120+185

Thus the value of N is 1120, and the sum of digits you can do yourself.

(Interestingly, we did not use the fact that the remainder had to be positive.)

What if, instead, we wanted the N with the largest sum of digits?

N still must divide each of 2240, 3360, and 5600, so our candidate set is the divisors of 1120.  Those are:

1, 2, 4, 5, 7, 8, 10, 14, 16, 20, 28, 32, 35, 40, 56, 70, 80, 112, 140, 160, 224, 280, 560, and 1120.

Of these, the N with the largest sum of digits is either 56 or 560.