Problem 1058

Solution. Let $A=\overline{a_{1} a_{2} \ldots a_{n}}$ and $B=\overline{b_{1} b_{2} \ldots b_{n}}$. From the condition $A+B=10^{n}$, it follows that the sum of the last two digits is 0 or 10.

If $a_{n}+b_{n}=10$, then from the condition $A+B=10^{n}$, it would follow that $a_{k}+b_{k}=9$ for every $k=1,2, \ldots, n-1$. By summing the equations, we get

$$a_{1}+a_{2}+\ldots a_{n}+b_{1}+b_{2}+\ldots+b_{n}=\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\ldots+\left(a_{n}+b_{n}\right)=9(n-1)+10$$

The left side of the equation is an even number because the digits of the number $A$ are the same as the digits of the number $B$. The right side of the equation is an odd number, since $9(n-1)$ is an odd number. In this way, we have a contradiction with the previous statement.

It follows that $a_{n}+b_{n}=0$, i.e., $a_{n}=0$ and $b_{n}=0$, since $a_{n}$ and $b_{n}$ are digits, which means that each of the numbers $A$ and $B$ is divisible by 10.

## 2nd year