Problem 10.8. A natural number $1 \leqslant n \leqslant 221$ is called lucky if the remainder of 221
divided by $n$ is divisible by the quotient (in this case, the remainder can be equal to 0). How many lucky numbers are there?
Answer: 115.

Solution. Let for some successful number $n$ the quotient be $k$, and the remainder be $k s$, by definition it is non-negative and less than the divisor $n$ (from the condition it follows that $k$ is a natural number, and $s$ is a non-negative integer.) Then

$$
221=n k+k s=k(n+s)
$$

Since $221=13 \cdot 17$ is divisible by the natural number $k$, we get several cases.

- Let $k=1$, then $n+s=221$. Since $0 \leqslant s110$. Notice that all such $n \in\{111,112,113, \ldots, 221\}$ are successful (and there are exactly 111 such numbers). Indeed, in all these cases $221=n \cdot 1+(221-n)$ the remainder (221-n) is non-negative, less than the corresponding $n$, and also divisible by the corresponding quotient 1.
- Let $k=13$, then $n+s=17$. Since $0 \leqslant 13 s14 s$, i.e., either $s=0$ and $n=17$, or $s=1$ and $n=16$. Notice that the numbers $n=16$ and $n=17$ are successful. Indeed, in both cases $221=16 \cdot 13+13$ and $221=17 \cdot 13+0$ the remainder is non-negative, less than the corresponding $n$, and also divisible by the corresponding quotient.
- Let $k=17$, then $n+s=13$. Since $0 \leqslant 17 s18 s$, i.e., $s=0$ and $n=13$. Notice that $n=13$ is successful. Indeed, in the case $221=$ $13 \cdot 17+0$ the remainder is non-negative, less than the corresponding $n$, and also divisible by the corresponding quotient.
- Let $k=221$, then $n+s=1$. Since $0 \leqslant 221 s222 s$, i.e., $s=0$ and $n=1$. Notice that $n=1$ is successful. Indeed, in the case $221=$ $1 \cdot 221+0$ the remainder is non-negative, less than the corresponding $n$, and also divisible by the corresponding quotient.

In total, there are $111+2+1+1=115$ successful numbers.

## 11th grade