3. (3 points) Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12.
Solution. Let $r_{n}(a)$ be the remainder of the division of the number $a$ by the number $n$. Suppose $a \in [1 ; 1000]$ and $r_{11}(a)=r_{12}(a)=t$. Then $t \in \{0, \ldots, 10\}$ and the following equality holds:

$$
t+11 k=t+12 m=a, \quad k, m \in N_{0}
$$

From the last equality, it follows that $k$ is divisible by 12, and $m$ is divisible by 11. Therefore,

$$
t+11 \cdot 12 \cdot s=t+132 \cdot s=a, \quad s \in N_{0}
$$

It remains to take into account the condition $a=t+132 \cdot s \leq 1000, \quad t \in \{0, \ldots, 10\}$:

$$
\begin{aligned}
& t=0 \Rightarrow 132 s \leq 1000 \Rightarrow s \leq 7.6 \Rightarrow 7 \text{ numbers } \quad(s \neq 0) \\
& t=1 \Rightarrow 1+132 s \leq 1000 \Rightarrow s \leq 7.57 \Rightarrow 8 \text{ numbers } \\
& t=2 \Rightarrow 2+132 s \leq 1000 \Rightarrow s \leq 7.56 \Rightarrow 8 \text{ numbers } \\
& t=3 \Rightarrow 3+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=4 \Rightarrow 4+132 s \leq 1000 \Rightarrow s \leq 7.55 \Rightarrow 8 \text{ numbers } \\
& t=5 \Rightarrow 5+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=6 \Rightarrow 6+132 s \leq 1000 \Rightarrow s \leq 7.54 \Rightarrow 8 \text{ numbers } \\
& t=7 \Rightarrow 7+132 s \leq 1000 \Rightarrow s \leq 7.53 \Rightarrow 8 \text{ numbers } \\
& t=8 \Rightarrow 8+132 s \leq 1000 \Rightarrow s \leq 7.52 \Rightarrow 8 \text{ numbers } \\
& t=9 \Rightarrow 9+132 s \leq 1000 \Rightarrow s \leq 7.51 \Rightarrow 8 \text{ numbers } \\
& t=10 \Rightarrow 10+132 s \leq 1000 \Rightarrow s \leq 7.5 \Rightarrow 8 \text{ numbers }
\end{aligned}
$$

In total, we get 87 numbers. (obviously, the last enumeration could have been shortened)

Answer: 87 numbers.

P.S. An alternative approach consists in enumerating $s$, where $s \leq \frac{1000}{132} \approx 7.57$, and finding those $t \in \{0, \ldots, 10\}$ for which the inequality $a=t+132 \cdot s \leq 1000$ holds.