2010 AMC 10A Problem 23

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it. The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white. This happens with a probability of $$\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$$ Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$ We have to find the smallest $n$ such that $$\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$$ $$n(n + 1) \gt 2010.$$ Guessing and checking gives us that the smallest $n$ that works is $45.$ Thus, A is the correct answer.