3. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400.
Solution. 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., $k=101 p, p \in \mathrm{Z}$. 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 $p=3 q, q \in \mathrm{Z}$, and the second when $p=3 q+1, q \in \mathrm{Z}$, from which we obtain that $k=303 q, k=303 q+101, q \in \mathrm{Z}$.

b) $(k-2): 101$, i.e., $k=101 p+2, p \in \mathrm{Z}$. Then we get $(101 p+2) 101 p:(3 \cdot 101) \Leftrightarrow(101 p+2) p \vdots 3$. The first factor is divisible by 3 when $p=3 q+2, q \in \mathrm{Z}$, and the second when $p=3 q, q \in \mathrm{Z}$, from which we obtain that $k=303 q+204$, $k=303 q+2, q \in \mathrm{Z}$.

Thus, the numbers that satisfy the condition of the problem are those that give remainders $0, 101, 204, 2$ when divided by 303, meaning that every 4 out of 303 consecutive numbers fit. Since $333300=303 \cdot 1100$, we get $4 \cdot 1100=4400$ numbers.