2016 AMC 10A Problem 17
Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that and that approaches as grows large. What is the sum of the digits of the least value of such that ?
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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 balls in the line, so the red ball can be in any of spots, each equally likely. If green balls are to its left, then can be , each with chance .
2. Try a small case, . We need at least greens on one side. That fails only when both sides have or fewer, which forces exactly. So bad spot out of , and .
3. In general write , so we need greens on one side. The rule fails when the left side has fewer than and the right side, , has fewer than . The second condition means is more than . So the bad values are
which is values. (For , , this gives bad spot, as found above.)
4. So
This matches the problem's checks: gives , and for large the fraction is close to , so is close to .
5. We want , which means the bad fraction is more than :
6. Cross-multiply (both denominators are positive): , so , which gives , or .
7. The smallest whole number is , so . (Check : is just under because .)
8. The digit sum of is .
The answer is .
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 ( and the limit ); 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 positions, and each position is equally likely. If green balls are to its left, then are to its right, and takes each value with probability .
Write . The condition fails when neither side holds at least greens, i.e. and , which means . The failing values are , a total of positions. Hence
(Check: gives , and as the fraction tends to , so .)
We need :
So the least is , whose digit sum is .
The answer is .
Review in the AoPS books
- Introduction to Counting & Probability, Chapter 7: Introduction to Probability
- Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
- Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)
- Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques
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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.