HMMT November, 2021

Number theory Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $n$ be the answer to this problem. An urn contains white and black balls. There are $n$ white balls and at least two balls of each color in the urn. Two balls are randomly drawn from the urn without replacement. Find the probability, in percent, that the first ball drawn is white and the second is black.

Solution

Solution:

Let the number of black balls in the urn be $k \geq 2$ . Then the probability of drawing a white ball first is $\frac{n}{n+k}$ , and the probability of drawing a black ball second is $\frac{k}{n+k-1}$ . This gives us the equation
$\frac{n k}{(n+k)(n+k-1)} = \frac{n}{100}$
from which we get
$(n+k)(n+k-1) = 100 k$
Let $m = n + k$ . Since $100 \mid m(m-1)$ , we must have that either $100$ divides one of $m, m-1$ or $25$ divides one of $m, m-1$ and $4$ divides the other. Since $m, m-1 > k$ , if either of $m$ or $m-1$ is greater than or equal to $100$ , the product $m(m-1) > 100 k$ . Therefore, the only possible values for $m$ are $25$ and $76$ .

If $m = 25$ , we have
$m(m-1) = 600 \Longrightarrow k = 6 \Longrightarrow n = 19$
If $m = 76$ , we have
$m(m-1) = 5700 \Longrightarrow k = 57 \Longrightarrow n = 19$
So $n = 19$ is the unique solution.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.