+4 Suppose that $A$, $B$, and $C$ are non-zero distinct digits less than $6$, and suppose we have ${AB_6}+{C_6}={C}0_6$ and ${AB_6}+{BA_6}={CC_6}$. Find the three-digit number ${ABC}$. (Interpret $AB_6$ as a base-6 number with digits $A$ and $B$, not as $A$ times $B$. The other expressions should be interpreted in this way as well).
+1725 Converting the first equation to base 10, we have 6A+B+C=6C. Simplifying, we have 6A+B=5C.
Converting the second equation to base 10, we have 6A+B+6B+A=6C+C. Simplifying, we have 7A+7B=7C. Dividing both sides by 7, we have A+B=C.
Substituting C=A+B into the first equation, we have 6A+B+A+B=6(A+B). Simplifying, we have 8A+2B=6A+6B. Combining like terms, we have 2A=4B. Dividing both sides by 2, we have A=2B.
Since A, B, and C are distinct digits less than 6, the only possible values for (A,B,C) are (2,1,3) and (4,2,6). However, C must be less than 6, so the only solution is ABC = 213.
