Faithful numbers

We'll show that the only such numbers are $1,2,3,4,5,6,8,12,24$ . The sum of these is $\fbox{65}$ .

First note that multiples of faithful numbers are faithful. Next note that the following numbers are faithful: $\begin{aligned} 2d+1 &= 1+2+(2d-2) \quad (d \ge 3) \\ 16 &= 1+5+10 \\ 10 &= 1+3+6 \\ \end{aligned}$ Let $n$ be an unfaithful number. Write $n = 2^k m$ where $m$ is odd. Then the first two above identities imply that $m \in \{ 1,3,5 \}$ and $k \in \{ 0,1,2,3 \}$ , which leads to twelve possibilities. Three of these, $n = 10,20,40,$ are out by the third identity. It's easy to check that the other nine are unfaithful. So we are done.