3. Find the number of natural numbers $k$, not exceeding 353500, such that $k^{2}+k$ is divisible by 505.
Answer: 2800.

Solution: By factoring the dividend and divisor, we get the condition $k(k+1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k+1)$ is divisible by 101. Let's consider two cases.

a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+1):(5 \cdot 101) \Leftrightarrow p(101 p+1): 5$. The first factor is divisible by 5 when $p=5 q, q \in \mathbb{Z}$, and the second factor is divisible by 5 when $p=5 q+4, q \in \mathbb{Z}$, from which we get $k=505 q, k=505 q+404, q \in \mathbb{Z}$.

b) $(k+1): 101$, i.e., $k=101 p+100, p \in \mathbb{Z}$. Then we get $(101 p+100)(101 p+101):(5 \cdot 101) \Leftrightarrow (101 p+100)(p+1): 5$. The first factor is divisible by 5 when $p=5 q, q \in \mathbb{Z}$, and the second factor is divisible by 5 when $p=5 q+4, q \in \mathbb{Z}$, from which we get $k=505 q+100, k=505 q+504, q \in \mathbb{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $0, 404, 100, 504$ when divided by 505, meaning that every 4 out of 505 consecutive numbers fit. Since $353500=505 \cdot 700$, we get $4 \cdot 700=2800$ numbers.

![](https://cdn.mathpix.com/cropped/2024_05_06_9d4c14e78dfd739c92cag-28.jpg?height=120&width=1071&top_left_y=2122&top_left_x=90)

Answer: $(8 ; 2)$.

Solution: Transform the equation of the system (add $64 x^{2}+64 y^{2}$ to both sides):

$$
\begin{gathered}
\left(x^{2}+y^{2}\right)^{2}+144+24 x^{2}+24 y^{2}=64 x^{2}+64 y^{2}+128 x y \Leftrightarrow\left(x^{2}+y^{2}+12\right)^{2}=(8 x+8 y)^{2} \Leftrightarrow \\
\Leftrightarrow\left[\begin{array}{l}
x^{2} + y^{2} + 12 = 8 x + 8 y, \\
x^{2} + y^{2} + 12 = -8 x - 8 y
\end{array} \Leftrightarrow \left[\begin{array}{l}
(x-4)^{2}+(y-4)^{2}=20, \\
(x+4)^{2}+(y+4)^{2}=20 .
\end{array}\right.\right.
\end{gathered}
$$

We get two circles of radius $2 \sqrt{5}$ with centers at points $(4 ; 4)$ and $(-4 ;-4)$.

The inequality in the system defines a half-plane. Consider the relative positions of each circle with the line $y=2 x-14$, which is the boundary of this half-plane.
a) $\left\{\begin{array}{l}(x-4)^{2}+(y-4)^{2}=20, \\ y=2 x-14\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-4)^{2}+(2 x-18)^{2}=20, \\ y=2 x-14\end{array} \Leftrightarrow\left\{\begin{array}{l}5 x^{2}-80 x+320=0, \\ y=2 x-14\end{array} \Leftrightarrow\left\{\begin{array}{l}x=8, \\ y=2 .\end{array}\right.\right.\right.\right.$

b) $\left\{\begin{array}{l}(x+4)^{2}+(y+4)^{2}=20, \\ y=2 x-14\end{array} \Leftrightarrow\left\{\begin{array}{l}(x+4)^{2}+(2 x-10)^{2}=20, \\ y=2 x-14\end{array} \Leftrightarrow\left\{\begin{array}{l}5 x^{2}-32 x+96=0, \\ y=2 x-14\end{array} \Leftrightarrow \varnothing\right.\right.\right.$.

In this case, the centers of the considered circles, the points $(4 ; 4)$ and $(-4 ;-4)$, do not lie in the half-plane, as their coordinates do not satisfy the inequality. Therefore, the second circle has no common points with the half-plane, while the first circle has a unique common point $(8 ; 2)$.