A three-digit number has a digit sum of 16. If in this number the digits in the hundreds and tens places are swapped, the number decreases by 360. If in the original number the digits in the tens and units places are swapped, the number increases by 54.

Find that three-digit number.

(L. Hozová)
Let the original three-digit number be $\overline{a b c}=100 a+10 b+c$. According to the first piece of information from the problem, we have

$$
a+b+c=16 \text {. }
$$

According to the second piece of information, $\overline{b a c}=\overline{a b c}-360$, thus

$$
\begin{aligned}
100 b+10 a+c & =100 a+10 b+c-360, \\
360 & =90 a-90 b, \\
4 & =a-b .
\end{aligned}
$$

According to the third piece of information, $\overline{a c b}=\overline{a b c}+54$, thus

$$
\begin{aligned}
100 a+10 c+b & =100 a+10 b+c+54, \\
9 c-9 b & =54, \\
c-b & =6 .
\end{aligned}
$$

If we express $a=b+4$ and $c=b+6$ from the second and third pieces of information, respectively, and substitute them into the first, we get $3 b+10=16$, thus $b=2$. Substituting this result into the previous expressions, we obtain $a=6$ and $c=8$. The original three-digit number was 628.

Evaluation. 2 points for each of the equations (2) and (3); 2 points for solving the system and determining the unknown number. A correct solution without further commentary is worth 1 point.

Notes. The differences between numbers formed by swapping two digits are always multiples of nine, with the corresponding multiple reflecting the positions of the swapped digits. For example, in the previous solution, we see $\overline{a b c}-\overline{b a c}=90(a-b)$ and $\overline{a b c}-\overline{a c b}=9(b-c)$, similarly $\overline{a b c}-\overline{c b a}=99(a-b)$. Such or similar reasoning before actually solving the problem allows for a quicker derivation of equations (2) and (3).

The second and third pieces of information from the problem can be visually represented as follows:

$$
\begin{array}{r}
a b c \\
-b a c \\
\hline 360
\end{array} \quad \begin{array}{r}
a c b \\
-a b c
\end{array}
$$

By comparing the highest place values, we see that the difference $a-b$ is 3 or 4, and the difference $c-b$ is 5 or 6 (the two possibilities for each difference correspond to whether or not there is a carry-over). These constraints, together with (1), yield a unique solution that can be found by systematically testing the possibilities. For example, the conditions $a-b=4$ and $a+b+c=16$ are satisfied by the numbers 952, 844, 736, and 628, of which only the last one also satisfies the constraint on the difference $c-b$.