Task 1. If to a two-digit number the digit 5 is added once at the beginning, and another time at the end, two different numbers are obtained whose difference is 252. What is that two-digit number?
Solution. Let the two-digit number be denoted by $\overline{x y}=10 x+y$. Then from the condition of the problem, we have two possibilities:
a) $\overline{5 x y}-\overline{x y 5}=252$, from which we get:

$$
\begin{array}{ll}
5 \cdot 100+x \cdot 10+y-(x \cdot 100+y \cdot 10+5)=252 & \Leftrightarrow \\
500+10 x+y-100 x-10 y-5=252 & \Leftrightarrow \\
243=90 x+9 y & \Leftrightarrow \\
27 \cdot 9=9(10 x+y) & \Leftrightarrow \\
27=10 x+y &
\end{array}
$$

Thus, the desired number is 27.

b) The second possibility is $\overline{x y 5}-\overline{5 x y}=252$, from which, with a procedure analogous to the previous case, we get that $10 x+y=83$.

Finally, the desired two-digit numbers are 27 and 83.