3. Find the number of natural numbers $k$, not exceeding 242400, such that $k^{2}+2 k$ is divisible by 303.
Answer: 3200.

Solution. By factoring the dividend and divisor, we get the condition $k(k+2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k+2)$ is divisible by 101. Let's consider two cases.

a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+2):(3 \cdot 101) \Leftrightarrow p(101 p+2) \vdots 3$. The first factor is divisible by 3 when $p=3 q, q \in \mathbb{Z}$, and the second when $p=3 q+2, q \in \mathbb{Z}$, from which we obtain that $k=303 q, k=303 q+202, q \in \mathbb{Z}$.

b) $(k+2): 101$, i.e., $k=101 p+99, p \in \mathbb{Z}$. Then we get $(101 p+99)(101 p+101):(3 \cdot 101) \Leftrightarrow (101 p+99)(p+1): 3$. The first factor is divisible by 3 when $p=3 q, q \in \mathbb{Z}$, and the second when $p=3 q+2, q \in \mathbb{Z}$, from which we obtain that $k=303 q+99, k=303 q+301, q \in \mathbb{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders 0, 202, 99, 301 when divided by 303, meaning that every 4 out of 303 consecutive numbers fit. Since $242400=303 \cdot 800$, we get $4 \cdot 800=3200$ numbers.