Ok, that means...

If N is a 4-digit number then

$\displaystyle 1000a+100b+10c+d=13(a+b+c+d)$

$\displaystyle \Rightarrow d = \frac{329a-29b-c}{4}$

Even if you have a = 1, b = 0, c = 9... d = 80. Any other values of $\displaystyle a\neq 0$, b, c only give larger values of d. I'm guessing this would only be exacerbated if N had > 4 digits.

So N must have 3 digits.

Hohoho, I think there's a good way of doing this.

As you said, $\displaystyle c=\frac{29a-b}{4}$

Then $\displaystyle 29a-b \equiv 0$ (mod 4), so $\displaystyle 29a \equiv b$ (mod 4). That is, $\displaystyle b\text{mod}4=1$. So $\displaystyle b=1,5,9$

3 Solutions (D)!