A number theory problem by Anca Baltariga

Clearly $y$ and $z$ cannot exceed $1$ , because $x + 2y + 3z = (x+2) + 2(y-1) + 3z = (x+3) + 2y + 3(z-1)$ . Thus $y=z=1$ . Also, $x$ does not exceed $2$ , because $x+2y + 3z = (x-2) + 2(y+1) + 3z$ . Thus $x+2y+3z = x+5 \leq 7$ , and it is easy to see that $7$ works.

If that number is greater than 7, it can be expressed as 3x1+2x1+n or 3x1+2x2+m (n and m are positive integers). If that number is 7, it can be expressed uniquely as 3x1+2x1+2 (1,1,2 are positive integers). Hence the result is 7.