Find a two-digit number that, when divided by the sum of its digits, gives one third of the sum of its digits.

(Bálintová)
The sought two-digit number can be written as $10a + b$, its digit sum would then be $a + b$. The problem statement can thus be rewritten as:

$$
(10a + b):(a + b) = \frac{a + b}{3}.
$$

This can be rewritten in the form

$$
10a + b = \frac{(a + b)^2}{3}.
$$

We see that the digit sum of the sought number must be divisible by three, so only these digit sums are possible: $3, 6, 9, 12, 15, 18$. Using the above relationship, we get the following numbers in sequence: $3, 12, 27, 48, 75, 108$. After performing the check, we find that only two numbers satisfy the problem statement, and these are 27 and 48.

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