52 (975). Find a two-digit number, the doubled sum of whose digits is equal to their product.
Solution. Let the desired two-digit number be $\overline{a b}=10 a+b$. By the condition $2(a+b)=a b$. To solve this equation, we express one of the unknowns (for example, $a$) in terms of the other ($b$). We will have:

$$
\begin{aligned}
& 2 a-a b=-2 b, a(2-b)=-2 b \\
& a=\frac{2 b}{b-2}=\frac{2(b-2)+4}{b-2}=2+\frac{4}{b-2}
\end{aligned}
$$

The number $b \neq 2$ (otherwise, from the equation $2(a+b)=a b$ it would follow that $4=0$). By trial, we find those values of $b \in\{1,3,4, \ldots, 9\}$ for which the fraction $\frac{4}{b-2}$ is an integer greater than -2 (otherwise $a<0$). We get $b_{1}=3, b_{2}=4, b_{3}=6$. From which $a_{1}=6$, $a_{2}=4, a_{3}=3$

Answer: $63,44,36$.

$53(976)$. Does there exist an ordinary fraction with a denominator of 20 that belongs to the interval $\left(\frac{4}{13} ; \frac{5}{13}\right)$?

Solution. Suppose $\frac{4}{13}<\frac{a}{20}<\frac{5}{13}$, where $a \in \boldsymbol{N}$. Bringing the fractions to a common denominator, we will have:

$$
\frac{4 \cdot 20}{13 \cdot 20}<\frac{a \cdot 13}{20 \cdot 13}<\frac{5 \cdot 20}{13 \cdot 20}
$$

from which we get the inequality $80<13 a<100$, which has a unique natural solution $a=7$. Thus, there exists a unique ordinary fraction with a denominator of 20, $\frac{7}{20}$, that belongs to the interval $\left(\frac{4}{13} ; \frac{5}{13}\right)$.

$\triangle 54$ (977). There are 6 locks and 6 keys to them. How many trials are needed to establish the correspondence between the keys and the locks?

Solution. The first key, in the worst case, will need to be tested 5 times (if it does not fit each of the five locks, it means it corresponds to the sixth). The second key, in the worst case, is tested 4 times, the third - 3 times, and so on. Thus, in the least favorable case, to establish the correspondence between 6 keys and 6 locks, 15 trials are required $(5+4+3+2+1=15)$.

Answer: 15 trials.