2016 AMC 10A Problem 17

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

Place the red ball in one of the

N+1

gaps among the green balls

The bad gaps have between

\frac{2N}{5}

and

\frac{3N}{5}

green balls on the left

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

\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

is

480

, whose digit sum is

12

.

Thus, the correct answer is A.

Problem 17 in other years