Sum of the digits

Question

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

Answer 1

Hint: One way to think of $a$ and $b$ as leaving the same remainder upon division by $m$ is $a\equiv b\pmod{m}$

Answer 2

Since division by $N$ leaves the same remainder for all three numbers, $N$ leaves zero remainder when divided into the difference of each pair.

Answer 3

$4665-1305=3360$ so the numbers need to divide $3360$
$6905-4665=2240$ so the numbers need to divide $2240$
The numbers need to divide $1120=2^5\cdot5\cdot7$
Knowing the prime factorization of a number, you can find the sum of all the factors of that number.

Hint 1: The sum of the factors of $p^n$ is $$ 1+p+p^2+p^3+\dots+p^n=\frac{p^{n+1}-1}{p-1} $$ Hint 2: We can find all the factors of $p^3q^2$ in the product $$ (1+p+p^2+p^3)(1+q+q^2) $$