Sum of possible all possible $x$ such that $51 \equiv 3 \pmod{x}$

Question

I was asked this simple following question:

What is the sum of all positive integers $x$ such that : $$51\equiv 3 \pmod{x}$$

My answer is $118$ (and I am pretty sure it's right but would like to check because there are other answers also),

My solution is this : $$51=kx+3 \Rightarrow 48=kx\quad $$ and $ x>3$ , therefore only $4,6,8,12,16,24,48$ are possible values for $x$ .

Have I missed anything?

Answer 1

Why the restriction that $x>3$? It also works when $x\in\{1,2,3\}.$

Aside from that, it looks fine.

Answer 2

$51\equiv 3 \pmod x$ iff $x|51-3=48$. Divisors of $48$ are $1,2,3,4,6,8,12,16,24,48$ .

Answer 3

$51 \equiv 3 \pmod x \iff 48 \equiv 0 \pmod x$

thus $x=1,2,3,4,6,8,12,16,24,48$ and their sum is

$S = 124$.