Question

Let $N$ be a positive multiple of $5.$ One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\frac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\frac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \dfrac{321}{400}?$

$12$

$14$

$16$

$18$

$20$

Solution:

Think of first arranging the $N$ green balls, then placing the red ball in one of the $N+1$ gaps. If $k$ green balls are to the left of the red ball, then $N-k$ are to its right.

At least $\frac35N$ green balls are on one side exactly when $k\le\frac25N$ or $k\ge\frac35N$ . Thus the bad gaps are $\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$ a total of $\frac15N-1$ gaps.

Therefore $P(N)=1-\frac{N/5-1}{N+1}=\frac{4N+10}{5N+5}.$ Solving $\frac{4N+10}{5N+5}<\frac{321}{400}$ gives $N>479$ . The least positive multiple of $5$ is $480$ , whose digit sum is $12$ .

Thus, the correct answer is A.

2016 AMC 10A Problem 17

Solution:

Problem 17 in Other Years