2. In the children's creativity center, there are 32 clubs. It is known that 6 children attend each club, and in any two clubs, the total number of children is 13. How many children in total attend the children's creativity center.

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# Answer: There are no solutions.

Solution (Attention to detail problem): Since six children attend each circle, the total number of children attending two circles does not exceed 12, while the problem states there are 13. Therefore, this is impossible.

## Recommendations for checking.

Only the answer - 0 points.

3. Solve the system of equations:

$\left\{\begin{array}{l}\sin x \cdot \cos y+\cos x=\sqrt{2} \\ x \cdot \cos x+y \cdot \sin y=\frac{\sqrt{2} \pi}{8}\end{array}\right.$

Answer: $x=\frac{\pi}{4}, y=2 \pi k, k \in Z$

Solution: Consider the first equation of the system $\sin x \cdot \cos y+\cos x=\sqrt{2} \Leftrightarrow \sqrt{1+\cos ^{2} y} \sin (x+\varphi)=\sqrt{2}$, where $\left\{\begin{array}{l}\cos \varphi=\frac{\cos y}{\sqrt{1+\cos ^{2} y}} \\ \sin \varphi=\frac{1}{\sqrt{1+\cos ^{2} y}}\end{array}\right.$

Since $1 \leq 1+\cos ^{2} y \leq 2,-1 \leq \sin (x+\varphi) \leq 1$, then

$\sqrt{1+\cos ^{2} y} \sin (x+\varphi)=\sqrt{2} \Leftrightarrow\left\{\begin{array}{c}\sqrt{1+\cos ^{2} y}=\sqrt{2} \\ \sin (x+\varphi)=1\end{array} \Leftrightarrow\left\{\begin{array}{c}\cos ^{2} y=1 \\ \sin (x+\varphi)=1\end{array} \Leftrightarrow\left\{\begin{array}{c}\sin y=0 \\ \sin (x+\varphi)=1\end{array}\right.\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}y=\pi n, n \in Z \\ \sin (x+\varphi)=1\end{array}\right.$. Therefore, $\varphi=\frac{\pi}{4}$ for even $n$ and $\varphi=\frac{3 \pi}{4}$ for odd $n$.

Then we get $\left\{\begin{array}{c}y=2 \pi k, k \in Z \\ x=\frac{\pi}{4}+2 \pi m, m \in Z\end{array}\right.$ or

$\left\{\begin{array}{c}y=\pi+2 \pi l, l \in \mathrm{Z} \\ x=-\frac{\pi}{4}+2 \pi p, p \in Z\end{array}\right.$. Substitute the obtained values into the second equation of the system and select only the solutions that satisfy it. The second equation of the system in the first case becomes:

$$
\left(\frac{\pi}{4}+2 \pi m\right) \cos \left(\frac{\pi}{4}+2 \pi m\right)=\frac{\sqrt{2} \pi}{8} \Leftrightarrow \frac{\pi}{4}+2 \pi m=\frac{\pi}{4} \Leftrightarrow m=0
$$

In the second case: $\left(-\frac{\pi}{4}+2 \pi p\right) \cos \left(-\frac{\pi}{4}+2 \pi p\right)=\frac{\sqrt{2} \pi}{8} \Leftrightarrow-\frac{\pi}{4}+2 \pi p=\frac{\pi}{4}-$ no solutions.

## Recommendations for checking.

First equation solved, but solutions satisfying the second equation not selected - 4 points.