Tolpygo A.K.

It is known that the number $a$ is positive, and the inequality $1 < x a < 2$ has exactly three solutions in integers. How many solutions in integers can the inequality $2 < x a < 3$ have?
On a half-interval of length 1 on the number line, there is always an integer, so on an interval of length greater than $n$ - there are at least $n$ integers. Therefore, if an interval contains exactly $n$ integers, then its length is greater than $n-1$ (the distance between the extreme integers), but not more than $n+1$ (otherwise, there would be $n+1$ integers).

The solutions of the inequalities lie on the intervals ( $\left(1 / a,{}^{2} / a\right)$ and $\left(2 / a,{}^{3} / a\right)$ of the same length. The first interval contains 3 numbers, so its length is greater than 2, but not more than 4. Therefore, the second interval contains no fewer than two, but no more than four integers. All these values are achieved, for example, when

$a=5 / 13,1 / 4,5 / 17$ respectively ( $1 / a=23 / 5,4,3^{2} / 5$ ).

## Answer

2, 3, or 4 solutions.