AMC 10 Step by Step

2016 AMC 10A Problem 17

P16-20~7 minPrint

Let NN be a positive multiple of 55 . One red ball and NN green balls are arranged in a line in random order. Let P(N)P(N) be the probability that at least 35\tfrac{3}{5} of the green balls are on the same side of the red ball. Observe that P(5)=1P(5)=1 and that P(N)P(N) approaches 45\tfrac{4}{5} as NN grows large. What is the sum of the digits of the least value of NN such that P(N)<321400P(N) < \tfrac{321}{400} ?

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Key insight

Only the red ball's spot matters; all N + 1 spots are equally likely. For N = 5m, exactly m - 1 middle spots fail.

Solution

The green balls all look alike, so an arrangement is decided by one thing: where the red ball sits.

1. There are N+1N + 1 balls in the line, so the red ball can be in any of N+1N + 1 spots, each equally likely. If LL green balls are to its left, then LL can be 0,1,2,…,N0, 1, 2, \ldots, N, each with chance 1N+1\frac{1}{N+1}.
2. Try a small case, N=10N = 10. We need at least 35⋅10=6\tfrac35 \cdot 10 = 6 greens on one side. That fails only when both sides have 55 or fewer, which forces L=5L = 5 exactly. So 11 bad spot out of 1111, and P(10)=1011P(10) = \tfrac{10}{11}.
3. In general write N=5mN = 5m, so we need 3m3m greens on one side. The rule fails when the left side has fewer than 3m3m and the right side, 5m−L5m - L, has fewer than 3m3m. The second condition means LL is more than 2m2m. So the bad values are
L=2m+1, 2m+2, …, 3m−1,L = 2m + 1,\ 2m + 2,\ \ldots,\ 3m - 1,
which is m−1m - 1 values. (For N=10N = 10, m=2m = 2, this gives 11 bad spot, as found above.)
4. So
P(N)=1−m−15m+1.P(N) = 1 - \frac{m - 1}{5m + 1}.
This matches the problem's checks: m=1m = 1 gives P=1P = 1, and for large mm the fraction is close to m5m=15\tfrac{m}{5m} = \tfrac15, so PP is close to 45\tfrac45.
5. We want P(N)<321400P(N) < \tfrac{321}{400}, which means the bad fraction is more than 1−321400=794001 - \tfrac{321}{400} = \tfrac{79}{400}:
m−15m+1>79400.\frac{m - 1}{5m + 1} > \frac{79}{400}.
6. Cross-multiply (both denominators are positive): 400(m−1)>79(5m+1)400(m - 1) > 79(5m + 1), so 400m−400>395m+79400m - 400 > 395m + 79, which gives 5m>4795m > 479, or m>95.8m > 95.8.
7. The smallest whole number mm is 9696, so N=5⋅96=480N = 5 \cdot 96 = 480. (Check m=95m = 95: 94476\tfrac{94}{476} is just under 79400\tfrac{79}{400} because 94⋅400=37600<79⋅476=3760494 \cdot 400 = 37600 < 79 \cdot 476 = 37604.)
8. The digit sum of 480480 is 4+8+0=124 + 8 + 0 = 12.

The answer is (A) 12\boxed{\textbf{(A)}\ 12}.

Why this works

A long line of identical balls with one special ball has a tiny set of outcomes: just the special ball's spot. The bad spots form one unbroken block in the middle, so counting them is easy, but it is easy to be off by one. The problem hands you two checks (P(5)=1P(5) = 1 and the limit 45\tfrac45); test your formula against them before solving.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The arrangement is determined by where the red ball sits among the N+1N+1 positions, and each position is equally likely. If LL green balls are to its left, then N−LN - L are to its right, and LL takes each value 0,1,…,N0, 1, \dots, N with probability 1N+1\frac{1}{N+1}.

Write N=5mN = 5m. The condition fails when neither side holds at least 3m3m greens, i.e. L<3mL < 3m and 5m−L<3m5m - L < 3m, which means 2m<L<3m2m < L < 3m. The failing values are L=2m+1,…,3m−1L = 2m+1, \dots, 3m-1, a total of m−1m - 1 positions. Hence
P(N)=1−m−15m+1.P(N) = 1 - \frac{m-1}{5m+1}.
(Check: m=1m = 1 gives P=1P = 1, and as m→∞m \to \infty the fraction tends to 15\frac15, so P→45P \to \frac45.)

We need m−15m+1>1−321400=79400\frac{m-1}{5m+1} > 1 - \frac{321}{400} = \frac{79}{400}:
400(m−1)>79(5m+1)  ⟺  400m−400>395m+79  ⟺  5m>479  ⟺  m≥96.400(m-1) > 79(5m+1) \iff 400m - 400 > 395m + 79 \iff 5m > 479 \iff m \ge 96.

So the least NN is 5⋅96=4805 \cdot 96 = 480, whose digit sum is 4+8+0=124 + 8 + 0 = 12.

The answer is (A) 12\boxed{\textbf{(A)}\ 12}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.