1. Determine all three-digit numbers less than 550 such that the digit in the hundreds place is equal to the product of the other two digits.
Solution. Let $\overline{a b c}$ be the desired number. Since $\overline{a b c}<550$, it follows that $a \in\{1,2,3,4,5\}$.

If $a=1$, then $b \cdot c=1$, from which it follows that $b=1$ and $c=1$, i.e., the desired number is 111.

If $a=2=1 \cdot 2=2 \cdot 1$, then the numbers obtained are 212 and 221.

If $a=3=1 \cdot 3=3 \cdot 1$, then the desired numbers are 313 and 331.

If $a=4=1 \cdot 4=4 \cdot 1=2 \cdot 2$, then the desired numbers are 414, 422, and 441.

If $a=5=1 \cdot 5=5 \cdot 1$, then the desired numbers are 515 and 551. Since the three-digit numbers must be less than 550, the number 551 is not a solution.

Finally, the desired numbers are: $111, 212, 221, 313, 331, 414, 441, 422$, and 515.