Problem 14 · 1991 AJHSME
Stretch
Counting & Probability
worst-caseparity
Several students are competing in a series of three races. A student earns 5 points for winning a race, 3 points for finishing second, and 1 point for finishing third. There are no ties. What is the smallest number of points a student must earn in the three races to be guaranteed of earning more points than any other student?
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Answer: D — 13.
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Hint 1 of 3
"Guaranteed" is the key word — you must beat the rival's BEST possible day, not their average. So picture the strongest single opponent and ask: what score do I need so even they can't tie me? Also: every place (5, 3, 1) is odd, so what's true of every three-race total?
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Hint 2 of 3
Three odd numbers always sum to an odd number, so all totals are odd: …, 9, 11, 13, … Test the candidates: for each possible score of yours, work out the best a rival could grab from the places you didn't take, and find the smallest score that leaves every rival strictly behind.
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Hint 3 of 3
If you score 11 (say 5+3+3), a rival can also reach 11, so 11 isn't safe. Bump to the next odd total and check whether any rival can still match it.
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Approach: beat the rival's best-case score, not their typical one
- "Guaranteed to win" means your total must exceed what the strongest possible rival could score, so think worst-case for you. First, a quick filter: 5, 3, 1 are all odd, and odd + odd + odd is always odd — so every three-race total is odd. Only 9, 11, 13, 15 are reachable; check from the bottom.
- Suppose you score 11. One way is 5 + 3 + 3 — but then a rival could take the two 1st places you didn't (5 + 5) plus a 1, reaching 11 and tying you. So 11 can be tied; not safe.
- Now score 13 — say 5 + 5 + 3 (win two races, 2nd in the third). The best any single rival can still collect is 2nd in your two wins and 1st in the remaining race: 3 + 3 + 5 = 11. That's strictly less than 13, every time.
- So the smallest guaranteed-winning total is 13.
- Why this transfers: "guarantee" problems are worst-case problems — you optimize against an adversary playing their best, not against an average. Pair that with a parity filter (totals here must be odd) to skip half the candidates instantly.
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