2016 AMC 10A Problem 17

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}

N

grows large. What is the sum of the digits of the least value of

N

such that

P(N) < \dfrac{321}{400}?

Show solution

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

k\ge\frac35N

. Thus the bad gaps are

\frac25N+1,\frac25N+2,\ldots,\frac35N-1,

a total of

\frac15N-1

gaps. Therefore

\begin{aligned} P(N) &= 1-\frac{\frac{N}{5}-1}{N+1} \\ &= \frac{4N+10}{5N+5}. \end{aligned}

Solving

\frac{4N+10}{5N+5}<\frac{321}{400}

gives

N>479

. The least positive multiple of

5

480

, whose digit sum is

12

. Thus, the correct answer is A.