# Task 7. (14 points)

In the analysis of bank accounts, it was found that the remaining balance on each of them is more than 10 rubles. It also turned out that there is a group of clients, each of whom has the same amount of money on their account. This amount is a number consisting of only ones. If you add up all the money on the accounts of this group of clients, the resulting sum will also be represented by a number consisting of only ones. Find the smallest number of clients in the group for which this is possible, if the group has more than one person.
# Solution

This problem is equivalent to the following.

Find the smallest natural number $m$, for which there exist natural numbers $n$ and $k$, such that $n>k>1$ and $\underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{k} \cdot m$.

Obviously, $m>9$. If $m=\overline{a b}$, where $a \geq 1$, then the equality $\underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{k} \cdot \overline{a b}$ means that $b=1$. However, in this case, regardless of $a$, the second digit from the right in the number $\underbrace{11 \ldots 1}_{k} \cdot \overline{a b}$ is $a+1$, and this cannot be 1. Therefore, $m \geq 100$. Clearly, $n=100$ does not satisfy the condition because $\underbrace{11 \ldots 1 \cdot 100}_{k}=\underbrace{11 \ldots 100}_{k}$.

On the other hand, $n=101$ works, since 101$\cdot$11=1111.

Answer: 101.