HMMT 十一月 2025 · 冲刺赛 · 第 25 题
HMMT November 2025 — Guts Round — Problem 25
题目详情
- [13] Jacob finished a rather hectic 9 holes of golf: he scored a 1 on the first hole, a 2 on the second hole, and so on, ending with a 9 on the ninth hole. Each hole is assigned a par of 3, 4, or 5. For each hole, Jacob subtracts its par from his score on that hole. Given that these 9 (possibly negative) differences can be arranged to form a sequence of 9 consecutive integers, compute the number of ways the pars could have been assigned to the holes.
解析
- [13] Jacob finished a rather hectic 9 holes of golf: he scored a 1 on the first hole, a 2 on the second hole, and so on, ending with a 9 on the ninth hole. Each hole is assigned a par of 3, 4, or 5. For each hole, Jacob subtracts its par from his score on that hole. Given that these 9 (possibly negative) differences can be arranged to form a sequence of 9 consecutive integers, compute the number of ways the pars could have been assigned to the holes. Proposed by: Sebastian Attlan Answer: 57 Solution: We do casework on the smallest difference, which is at least 1 − 5 = − 4 and at most 1 − 3 = − 2. • If the smallest difference is − 4, then the differences must be − 4 , − 3 , − 2 , . . . , 4, forcing all the pars to be 5. • If the smallest difference is − 2, then the differences must be − 2 , − 1 , 0 , . . . , 6, forcing all the pars to be 3. • If the smallest difference is − 3, then the differences must be − 3 , − 2 , − 1 , . . . , 5. Let f ( n ) de- note the number of ways to choose the pars for holes 1 , 2 , . . . , n such that the differences are − 3 , − 2 , − 1 , . . . , n − 4. We have f (1) = 1 (we must set the par to 4) and f (2) = 2 (we can set the pars to 3 , 5 or 4 , 4). We wish to compute f (9). We derive a recursion for f ( n ). Assume n ≥ 3. – The par of the n th hole cannot be 3 because that would result in a difference of n − 3. – If the par of the n th hole is 4, then we need to choose the pars for holes 1 , 2 , . . . , n − 1 such that the differences are − 3 , − 2 , − 1 , . . . , n − 5. There are f ( n − 1) ways to do so. – If the par of the n th hole is 5, then consider the ( n − 1)th hole. If it has a par of 4, then both the ( n − 1)th and n th differences would be n − 4, which is impossible. If it has a par of 5, then one of the first n − 2 differences must be n − 4, which is also impossible. Therefore, the ( n − 1)th hole must have a par of 3, meaning the ( n − 1)th and n th differences are n − 4 and n − 5. We now need to choose the pars for the holes 1 , 2 , . . . , n − 2 such that the differences are − 3 , − 2 , . . . , n − 6. There are f ( n − 2) ways to do so. Therefore, f ( n ) satisfies the Fibonacci recursion. We can compute f (3) = 3 , f (4) = 5 , f (5) = 8 , f (6) = 13 , f (7) = 21 , f (8) = 34 , f (9) = 55 . Hence, there are 55 possibilities in this case. Finally, the answer is 1 + 1 + 55 = 57 .