2010 AMC 10A Problem 23

Start with small cases to see the pattern.

1. What is in each box. Box $k$ has $1$ red and $k$ white marbles, so $k + 1$ marbles. The chance of white is $\dfrac{k}{k+1}$, and the chance of red is $\dfrac{1}{k+1}$.
2. Stopping after exactly $n$ draws means: white from box $1$, white from box $2$, and so on up to box $n - 1$, and then red from box $n$. Multiply these chances.
3. Small cases.
- $n = 1$: red from box 1: $\tfrac12$.
- $n = 2$: white from box 1, red from box 2: $\tfrac12 \times \tfrac13 = \tfrac16$.
- $n = 3$: $\tfrac12 \times \tfrac23 \times \tfrac14 = \tfrac{2}{24} = \tfrac1{12}$.
4. The pattern. In general,
$$P(n) = \frac12 \times \frac23 \times \frac34 \times \cdots \times \frac{n-1}{n} \times \frac{1}{n+1}.$$
Each top number cancels the bottom number just before it, so the white part shrinks to $\tfrac1n$. So
$$P(n) = \frac{1}{n} \times \frac{1}{n+1} = \frac{1}{n(n+1)}.$$
Check: $n = 3$ gives $\tfrac{1}{3 \times 4} = \tfrac1{12}$. It matches.
5. Solve the inequality. $P(n) < \dfrac{1}{2010}$ means $n(n+1) > 2010$. Now $44 \times 45 = 1980$, which is too small, and $45 \times 46 = 2070$, which is big enough.

The smallest $n$ is $45$.

The answer is $\boxed{\textbf{(A)}\ 45}$.

"Stops at step $n$" always means "keeps going at every step before $n$, then stops at $n$," so you multiply the chances. Fractions like $\tfrac12 \times \tfrac23 \times \tfrac34$ cancel in a chain, leaving a simple answer. Trying $n = 1, 2, 3$ first shows the pattern and gives a check.