Task A-1.2. (4 points)

Determine the natural number whose 9-fold is between 1100 and 1200, and whose 13-fold is between 1500 and 1600.

For a natural number $n$, the $n$-fold of a number is the number that is $n$ times larger than that number.
## Solution.

Let $n$ be the desired number. The conditions of the problem can be written as two inequalities:

$$
1100<9 n<1200, \quad 1500<13 n<1600
$$

By dividing the first inequality by 9 and the second by 13, we get:

$$
\frac{1100}{9}<n<\frac{1200}{9}, \quad \frac{1500}{13}<n<\frac{1600}{13}
$$

Since $n$ is a natural number, we have: $\quad 122<n \leqslant 133, \quad 115<n \leqslant 123$.

The only number that satisfies both inequalities is $n=123$.

Notes. It should also be recognized if the given conditions are interpreted as $1100 \leqslant 9 n \leqslant 1200$, $1500 \leqslant 13 n \leqslant 1600$, etc. A guessed solution (without justification) earns 1 point.