$a+b+c=9k,\,a≠b≠c,\,a,b,c∈N \, and\, a,b,c<10$

Question

how do you manage the following? \begin{equation} a+b+c=9k,a≠b≠c,a,b,c∈N \, and\, a,b,c<10 \end{equation} Prove that one of the adders must be multiple of three.

Thank you in advance.

Answer 1

Suppose we had such a triple $(a,b,c)$. Suppose, wlog, that $a<b<c$

First work $\pmod 3$. We have $a+b+c\equiv 0 \pmod 3$. If $a,b,c$ are distinct $\pmod 3$ then one of them is divisible by $3$. Thus two, at least, must coincide $\pmod 3$. Casework shows that we must have all three the same $\pmod 3$, so either we have $a,b,c\equiv 1 \pmod 3$ or we have $a,b,c\equiv -1\pmod 3$.

Say we have $a,b,c\equiv 1\pmod 3$. Then we must have $a,b,c\in \{1,4,7\}$. But $1+4+7=12$ is not a multiple of $9$.

Say we have $a,b,c\equiv -1 \pmod 3$. Then we must have $a,b,c\in \{2,5,8\}$. But $2+5+8=15$ is not a multiple of $9$.

And we are done.