Yashchenko I.v.

In all entrances of the house, there is the same number of floors, and on each floor, there is the same number of apartments. The number of floors in the house is greater than the number of apartments per floor, the number of apartments per floor is greater than the number of entrances, and the number of entrances is greater than one. How many floors are there in the house if there are a total of 105 apartments?
$105=3 \cdot 5 \cdot 7$.

## Solution

Let $p$ denote the number of entrances in the building, $f$ the number of floors, and $k$ the number of apartments per floor. Then $p \cdot f \cdot k=105=3 \cdot 5 \cdot 7$, where the numbers $3,5,7-$ are prime. Considering that $1<p<k<f$, we get $f=7$.

## Answer

7 floors.