Problem 23 · 2023 Math Kangaroo
Stretch
Counting & Probability
caseworkfactorization
13 athletes took part in a three-part climbing competition. There are no draws in any part. The final rank of each athlete is determined by arranging the products of the ranks in each of the three parts: if an athlete for example comes 4th once, 3rd once and 6th once, he has \(4 \cdot 3 \cdot 6 = 72\) points. The higher the number of points, the worse the final rank. What is the worst possible final rank Hans can get to if he was 1st in two of the parts?
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Answer: B — 3.
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Hint 1 of 3
Hans being 1st in two parts makes his score just his rank \(r\) in the third part, so a large \(r\) gives others more room.
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Hint 2 of 3
Other athletes never get rank 1 in the two parts Hans won, so their two factors there are at least 2 each.
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Hint 3 of 3
Count how many athletes can be arranged to have a product strictly below Hans's score.
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Approach: let Hans take the worst third-part rank, then squeeze others below him
- With two firsts Hans scores \(1 \cdot 1 \cdot r = r\); pick the largest helpful \(r = 10\) to leave room beneath him.
- Every other athlete has factors \(\ge 2\) in the two parts Hans won, so the smallest products come from athletes using the ranks 2 and 3 there: e.g. \(2\cdot 2\cdot 2 = 8\) and \(3\cdot 3\cdot 1 = 9\), both below 10, while no third athlete can be pushed under 10.
- So at most two athletes beat him, putting Hans at worst in final rank 3.
Mark:
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