Problem 6.4. Find any solution to the puzzle

$$
\begin{array}{r}
\mathrm{ABA} \\
+\mathrm{ABC} \\
\mathrm{ACC} \\
\hline 1416
\end{array}
$$

where $A, B, C$ are three different non-zero digits.

Enter the values of the digits $A, B, C$.

Instead of the letter $A$ should stand the digit:

Instead of the letter $B$ should stand the digit:

Instead of the letter $C$ should stand the digit:
Answer: $A=4, B=7, C=6$.

Solution. Suppose $A \leqslant 3$. Then $1416=\overline{A B A}+\overline{A B C}+\overline{A C C}<400+400+400=1200$ - contradiction.

Suppose $A \geqslant 5$. Then $1416=\overline{A B A}+\overline{A B C}+\overline{A C C} \geqslant 500+500+500=1500-$ contradiction.

Therefore, $A=4$. Considering the units digit, we get $A+2 C=4+2 C$ ends in 6, which is only possible if $C=1$ or $C=6$.

If $C=1$, then $1416=\overline{A B A}+\overline{A B C}+\overline{A C C} \leqslant 494+491+411=1396-$ contradiction. Therefore, $C=6$. Then

$$
1416=\overline{4 B 4}+\overline{4 B 6}+466=(404+\overline{B 0})+(406+\overline{B 0})+466=1276+2 \cdot \overline{B 0}
$$

from which we get $B=7$.