3. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505.
Answer: 3600.

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

b) $(k-1): 101$, i.e., $k=101 p+1, p \in \mathbb{Z}$. Then we get $(101 p+1) 101 p:(5 \cdot 101) \Leftrightarrow (101 p+1) p: 5$. The first factor is divisible by 5 when $p=5 q+4, q \in \mathbb{Z}$, and the second when $p=5 q, q \in \mathbb{Z}$, from which we obtain that $k=505 q+405$, $k=505 q+1, q \in \mathbb{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $0, 101, 405, 1$ when divided by 505, meaning that every 4 out of 505 consecutive numbers fit. Since $454500=505 \cdot 900$, we get $4 \cdot 900=3600$ numbers.