(I swear this isn't a trap problem)
Mary and her brothers would dispute a game called "all or one" to decide who would do the dishes.
The game works simply like this, its played by at least three people, and in each round each one choose a natural number simultaneously. We have a winner if, and only if, in certain round the person who won chose a certain number and all the others chose the same another one. Example: you choose the number 2 and the other 4 people (for example) choose the number 8, then you won. There is no limit of participants neither of the quantity of numbers available to be chosen.
And so, Mary chose the quantity $\mathbf{k}>1$ of numbers that are available to be chosen. To facilitate the understanding of her brothers about how the game works, she decided to choose this quantity in such way that the chance of some (fixed) person win in the first round is an integer percentage (example: 1%, 27%).
INTERESTING QUESTION: Knowing that $\mathbf{k}$ and the number of players in the game $\mathbf{n}$ (she and her brothers) are relatively primes, how many brothers does she have?
Assuming that $k$ is not permitted to be $1$, the solution is $n = 3$. Here's the reason:
The probability of the first $n - 1$ people all picking the same number is $\left(\frac{1}{k}\right)^{n - 2}$. The probability of the final person not picking that number is $\frac{k-1}{k}$. That means that the probability of the final person winning is $\left(\frac{1}{k}\right)^{n - 2}\frac{k-1}{k} = \frac{k-1}{k^{n-1}}$. Note that $k - 1$ and $k$ never have any nontrivial common factors, so this fraction is already in lowest form. If this is supposed to be an integer percentage, $\frac{k-1}{k^{n-1}} = \frac{m}{100}$ for some $m$. Since the left-hand fraction is in lowest form, its denominator must divide $100$. So $k^{n-1}$ divides $100$, and therefore $k^{n-1}$ is one of $1,2,4,5,10,20,25,50,$ or $100$.
$k$ isn't allowed to be $1$, so that kicks out the first option. We can rule out a lot of other things because $n$ is required to be at least $3$; that leaves only $4$, $25$, and $100$ as our options for $k^{n-1}$. In all three cases, $n$ can only be $3$. So $n = 3$ (and $k$ is either $2$, $5$, or $10$).