Evochkimov M.A.

Vasya thought of a two-digit number and told Petya the product of the digits in the number, and Sasha the sum of these digits. The boys had the following conversation:

Petya: "I can guess the number in three attempts, but two might not be enough".

Sasha: "If so, then I can guess it in four attempts, but three might not be enough".

What number was told to Sasha?
Let $\overline{a b}$ be the two-digit number in question, then $P=a b$ is the product of its digits. Since Petya undertakes to guess the number in three attempts, the number $P$ must be factorizable into two factors, which are digits, in exactly three ways, considering the order of the factors. Therefore:

1) $b \neq 0$ (otherwise $P=0=a \cdot 0$ for any $a$ from 1 to 9, which means there are 9 such ways);
2) For the numbers $\overline{a b}$ and $\overline{b a}$, the value of $P$ is the same, so $P$ must be the square of one of the digits (otherwise $P$ can be factorized in an even number of ways).

Thus, it is sufficient to consider the squares of all digits and list all ways to factorize them into the specified factors:

$\mathbf{1}=1 \cdot 1 ; \mathbf{4}=2 \cdot 2=1 \cdot 4=4 \cdot 1, \mathbf{9}=3 \cdot 3=1 \cdot 9=9 \cdot 1, \mathbf{1 6}=4 \cdot 4=2 \cdot 8=8 \cdot 2, \mathbf{2 5}=5 \cdot 5, \mathbf{3 6}=6 \cdot 6=4 \cdot 9=9 \cdot 4, \mathbf{4 9}=7 \cdot 7, \mathbf{6 4}=8 \cdot 8 ; 81=9 \cdot 9$.

Therefore, from Petya's statement, Sasha can conclude that the number in question could only be one of the following: 22, 14, 41, 33, 19, 91, 44, 28, 82, 66, 49, 94. Among them: one number has a digit sum of 4 (22), two numbers have a digit sum of 5 (14 and 41), one number has a digit sum of 6 (33), one number has a digit sum of 8 (44), four numbers have a digit sum of 10 (19, 91, 28, 82), one number has a digit sum of 12 (66), and two numbers have a digit sum of 13 (49 and 94). Since Sasha needs four attempts, the required sum is 10.

## Answer

10.