## Task A-1.1.

The competition takes place in 11 classrooms, each containing the same number of desks arranged in the same way: in a certain number of columns and a certain number of rows. There is one student in each desk. If there were one row of desks less and one column of desks more in each classroom, 10 classrooms would be sufficient, and there would still be two desks left empty. How many students can there be in total at the competition if it is known that their number is a three-digit number?
## Solution.

Let $N$ be the number of students in the state school, and $r$ and $s$ be the number of rows and columns of desks in each classroom.

In the first case, we have $N=11 r s$.

In the second case, we have $N+2=10(r-1)(s+1)=10 r s-10 s+10 r+10$.

By equating and rearranging, we get

$$
\begin{gathered}
11 r s+2=10 r s-10 s+10 r-10 \\
r s+12+10 s-10 r=0 \\
r(s-10)+10 s-100+100+12=0 \\
(r+10)(s-10)+112=0 \\
(10-s)(r+10)=112
\end{gathered}
$$

We consider all possible factorizations of the number 112 into two natural factors, where the first is less than 10 and the second is greater than 10. The divisors of 112 are $1,2,4,7,8,14,16,28,56,112$, so only the following factorizations are relevant:

- $10-s=1,10+r=112$, from which $s=9, r=102$, so $N=11 \cdot 9 \cdot 102=10098$;
- $10-s=2,10+r=56$, from which $s=8, r=46$, so $N=11 \cdot 8 \cdot 46=4048$;
- $10-s=4,10+r=28$, from which $s=6, r=18$, so $N=11 \cdot 6 \cdot 18=1188$;
- $10-s=7,10+r=16$, from which $s=3, r=6$, so $N=11 \cdot 3 \cdot 6=198$;
- $10-s=8,10+r=14$, from which $s=2, r=4$, so $N=11 \cdot 2 \cdot 4=88$.

From the conditions of the problem, $N$ is a three-digit number, so the only possible solution is $N=198$.