5. Solve the rebus:

$$
\Pi E T \cdot \Pi E T=Д E C E T
$$

(the same letters correspond to the same and different letters to different digits, and at the same time $\Pi, Д \neq 0)$.

## Second Grade - B Category
5. Since $400 \cdot 400 = 160000 > 99999$, we conclude $\Pi < 4$. Since $T \cdot T$ ends with $T$, it follows that $T \in \{0, 1, 5, 6\}$. We exclude the possibility $T = 0$, because then $PET^2$ would end with 00, which is not possible due to the right side.

Notice that the number $ET \cdot ET$ ends with $ET$, and since we have $ET^2 = (10 \cdot E + T)^2 = 100 \cdot E^2 + 20 \cdot E \cdot T + T^2$, we conclude that $20 \cdot E \cdot T + T^2$ must end with $ET$. Now we separate cases by $T$.

- $T = 1$: Then $20 \cdot E + 1$ must end with $E1$, i.e., $2 \cdot E$ must end with $E$, which is only possible for $E = 0$. Testing the possibilities $\Pi \in \{2, 3\}$, we get $201 \cdot 201 = 40401$, which is not a solution (due to $D \neq C$), and $301 \cdot 301 = 90601$, which is a solution.
- $T = 5$: Then $100 \cdot E + 25$ must end with $E5$, so we immediately get $E = 2$. Testing the possibilities $\Pi \in \{1, 3\}$, we get $125 \cdot 125 = 15625$ and $325 \cdot 325 = 105625$, and none of this is a solution.
- $T = 6$: Then $120 \cdot E + 36$ must end with $E6$, i.e., $2 \cdot E + 3$ must end with $E$, which is only possible for $E = 7$. Testing the possibilities $\Pi \in \{1, 2, 3\}$, we get $176 \cdot 176 = 30976$, $276 \cdot 276 = 76176$, and $376 \cdot 376 = 141376$, and none of this is a solution.

Therefore, the only solution is $301 \cdot 301 = 90601$.

## Second Grade - B Category