2. Find all three-digit numbers with this property: if we strike out the middle digit of the number and multiply the resulting two-digit number by the square of the struck-out digit, we get the original three-digit number again.

(Tomáš Jurík)
2. A three-digit number $n=\overline{a b c}$ has the desired property if and only if its digits $a$, $b$, $c$ satisfy the equation

$$
100 a+10 b+c=(10 a+c) b^{2} .
$$

Since the digit $b$ has the most complex representation in it, we will examine its possible values one by one.

First, we notice that $b \in\{0,1\}$ cannot be (for such digits, the right side of (1) would not be a three-digit number). Since $a \geqq 1$, $b \in\{5,6,7,8,9\}$ is also not possible - for such digits, the right side would be at least $250 a$, while the left side of (1) is always less than $100 a+100 \leqq 200 a$. Therefore, we need to examine the values $b \in\{2,3,4\}$.

For $b=2$, (1) becomes the equation $100 a+20+c=40 a+4 c$ or $20=3(c-20 a)$, which is impossible, as 20 is not a multiple of three (also, $c-20 a<0$ due to $a \geqq 1$ and $c \leqq 9$).

For $b=3$, (1) becomes the equation $100 a+30+c=90 a+9 c$ or $5(a+3)=4 c$, which means that $c$ is a non-zero digit divisible by five, i.e., $c=5$, and thus $a=1$. We have found the solution $n=135$ (verification is not necessary).

For $b=4$, (1) becomes the equation $100 a+40+c=160 a+16 c$ or $40=3(20 a+5 c)$, which is impossible, as 40 is not a multiple of three (also, $3(20 a+5 c) \geqq 60$).

Answer: The only number that satisfies the condition is 135.

For a complete solution, award 6 points, of which 1 point for setting up the equation, 1 point for excluding $b \leqq 1$, 1 point for excluding $b \geqq 5$, 1 point for excluding $b=2$, 1 point for excluding $b=4$, 1 point for solving $b=3$ and finding the number 135. In the case of an incomplete solution, award 1 point for guessing the solution.