5. The angle at vertex $B$ of triangle $A B C$ is $130^{\circ}$. Through points $A$ and $C$, lines perpendicular to line $A C$ are drawn and intersect the circumcircle of triangle $A B C$ at points $E$ and $D$. Find the acute angle between the diagonals of the quadrilateral with vertices at points $A, C, D$ and $E$.

Problem 1 Answer: 12 students.
Solution. Let $a$ be the number of students in the first category, $c$ be the number of students in the third category, and $b$ be the part of students from the second category who will definitely lie in response to the first question (and say "YES" to all three questions), while the rest of the students from this category will answer "NO" to all three questions. Then, "YES" to the first question will be given by $a+b+c=25$ students. "YES" to the second question will be given by $b+c=21$ students. "YES" to the third question will be given by $b=6$ students. Solving the system, we get: $a=4, b=6, c=15$. Then, to the second category, we should assign $31-a-c=12$ students.

Problem 2 Answer: $a= \pm 5-2 \sqrt{6}$.

Solution. By Vieta's theorem, the roots of the equation are $m=(a+2 \sqrt{6})$ and $n=\left(\frac{1}{a}-2 \sqrt{6}\right)$.

Then we have:

$$
\left[\begin{array}{l}
a=m-2 \sqrt{6} \\
\frac{1}{a}=n+2 \sqrt{6}
\end{array} \rightarrow 1=m n+2 \sqrt{6}(m-n)-24 \rightarrow 2 \sqrt{6}(m-n)=25-m n\right.
$$

If $m, n \in Z, m \neq n$, then $2 \sqrt{6}=\frac{25-m n}{m-n}$ is a rational number, which is incorrect. Therefore, $m=n$ and then $m^{2}=25 \rightarrow m= \pm 5 \rightarrow a= \pm 5-2 \sqrt{6}$.

Problem 3 Answer: 5400 numbers.

Solution. Write $496125=3^{4} \cdot 5^{3} \cdot 7^{2}$. There are a total of $3^{4} \cdot 5^{3}=10125$ multiples of 49. Among them, $3^{3} \cdot 5^{3}=3375$ numbers are divisible by 3, $3^{4} \cdot 5^{2}=2025$ numbers are divisible by 5, and $3^{3} \cdot 5^{2}=675$ numbers are divisible by 15. Then, among the numbers that are multiples of 49, there are $3375+2025-675=4725$ numbers that are divisible by either 3 or 5. Therefore, the numbers that are not divisible by either 3 or 5 will be $10125-4775=$ 5400 desired numbers.

Problem 4 Answer: $n=210$.

Solution. According to the problem, $n=7 k$ and the expression

$$
n^{2}+25 n+100=(n+5)(n+20)=(7 k+5)(7 k+20)
$$

must be divisible by 5 and 23. Note that if one of the factors is divisible by 5, then the other is also divisible by 5 and vice versa, so:

$$
7 k+5=5 m \rightarrow 7 k=5(m-1) \rightarrow\left\{\begin{array}{c}
k=5 t \\
m=7 t+1
\end{array} \rightarrow n=35 t, t \in Z\right.
$$

Case 1. $7 k+5$ is divisible by 23:

$$
35 t+5=23 u \rightarrow\left\{\begin{array}{l}
t=23 v-10 \\
u=35 v-15
\end{array} \rightarrow n=35 t=35(23 v-10) \rightarrow n_{\min }=455\right.
$$

Case 2. $7 k+20$ is divisible by 23:

$$
35 t+20=23 u \rightarrow\left\{\begin{array}{c}
t=23 v+6 \\
u=35 v+10
\end{array} \rightarrow n=35 t=35(23 v+6) \rightarrow n_{\min }=210\right.
$$

Choosing the smallest of the found $n_{\min }$, we get $n_{\min }=210$.

Problem 5 Answer: $80^{\circ}$.

Solution. Figures 1 and 2 show the possible geometric configurations.

![](https://cdn.mathpix.com/cropped/2024_05_06_e40112a47ee5f2126dcbg-2.jpg?height=409&width=440&top_left_y=1843&top_left_x=428)

Fig 1

![](https://cdn.mathpix.com/cropped/2024_05_06_e40112a47ee5f2126dcbg-2.jpg?height=349&width=342&top_left_y=1890&top_left_x=1234)

Fig 2

In the figures, point $O$ is the center of the circle $K$ circumscribed around triangle $A B C$.

Case 1. $\alpha>90^{\circ}$ (Fig. 1). Quadrilateral $A B C D$ is inscribed in circle $K$ by construction, so point $O$ is equidistant from points $C$ and $D$ at a distance equal to the radius of circle $K$. Similarly, point $O$ is equidistant from points $A$ and $E$ at the same distance. Therefore, point $O$ is the intersection point of the diagonals of rectangle $A C D E$. The angle $C D O$, as the opposite angle to $\alpha$ in the inscribed quadrilateral $A B C D$, is equal to $180^{\circ}-\alpha$. The desired angle between the diagonals of rectangle $A C D E$ is: $180^{\circ}-2\left(180^{\circ}-\alpha\right)=2\left(\alpha-90^{\circ}\right)$.

Case 2. $\alpha<90^{\circ}$ (Fig. 2). Point $O$ is equidistant from points $A, C, D$ and $E$, so it is the intersection point of the diagonals of rectangle $A C D E$ with these vertices. The angle $A D C$ is equal to $\alpha$, as it is inscribed and subtends the same arc of circle $K$ as angle $A B C$. Then the desired angle $D O C$ is: $180^{\circ}-2 \alpha=2\left(90^{\circ}-\alpha\right)$.

In case 1, $\alpha=130^{\circ}$, so case 1 applies, and $2\left(\alpha-90^{\circ}\right)=80^{\circ}$.