HMMT 二月 2026 · 冲刺赛 · 第 35 题
HMMT February 2026 — Guts Round — Problem 35
题目详情
- [20] Estimate ( ) 15000 ∑ 30000! log . 10 2 ( k !) (30000 − 2 k )! k =0 2 − 0 . 69( E − A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points.
解析
- [20] Estimate ( ) 15000 ∑ 30000! log . 10 2 ( k !) (30000 − 2 k )! k =0 2 − 0 . 69( E − A ) Submit a real number E . If the correct answer is A , you will receive ⌊ 20 . 05 e ⌋ points. Proposed by: Pitchayut Saengrungkongka Answer: 14311.088033944017 Solution: When 10000 is replaced by n , we claim that the exact asymptotic formula is ( ) 3 n/ 2 3 n ∑ 3 n 3 √ ∼ . k, k, 3 n − 2 k 2 πn k =0 (Recall the standard notation that f ( n ) ∼ g ( n ) if and only if lim f ( n ) /g ( n ) = 1 .) n →∞ Taking log of the right hand side gives a very accurate estimate of 10 ( √ ) 30000 log 3 − 2 − log 2 π = 14311 . 0880367 , 10 10 − 6 which is off by 2 . 8 · 10 , more than enough to get 20 points. ( ) 3 n Before we prove the asymptotic formula, we show how to derive it heuristically. Let f ( k ) = . k,k, 3 n − 2 k It is straightforward to show that f ( k ) is maximized at k = n (though we won’t use this in the actual estimation). Thus, we try to estimate the value of f ( k ) relative to f ( n ) . To that end, we note that if x ≪ n , then ( ) 2 2 x 1 − f ( n + x ) ( n − 2 x + 2)( n − 2 x + 1) 6 x − 6 x/n n = ≈ = 1 − ≈ e . ( ) 2 2 x f ( n + x − 1) ( n + x ) n 1 + n Thus, we have that ( ) f ( n + x ) 6 2 = exp − (1 + 2 + · · · + x ) ≈ exp( − 3 x /n ) , f ( n ) n so we have that the sum is approximately √ 3 n/ 2 ∫ ∞ ∞ ∑ ∑ πn 2 2 f ( k ) ≈ exp( − 3 x /n ) f ( n ) ≈ exp( − 3 x /n ) f ( n ) dx = f ( n ) , 3 −∞ x = −∞ k =0 ∫ √ 2 ∞ − x where the integral can be easily evaluated using Gaussian integral e dx = π . Finally, since −∞ √ ( ) n (3 n )! n f ( n ) = , we use Stirling’s formula n ! ∼ 2 πn to estimate f ( n ) , getting that the sum is 3 n ! e ( ) √ √ 3 n/ 2 3 n 3 n 1 / 2 3 n ∑ (2 π · 3 n ) πn πn 3 e √ f ( k ) ∼ f ( n ) ∼ · ∼ . ( ) 3 n n 3 / 2 3 3 2 πn (2 πn ) k =0 e To prove this rigorously, we need to keep track of error terms. If x ≤ n , then we have the estimate ( ) 2 x/n x x e = 1 + + O , with the constant in the error term not depending on n or x . Using this, we 2 n n get that f ( n + x ) ( n − 2 x + 2)( n − 2 x + 1) = 2 f ( n + x − 1) ( n + x ) ( ( )) 2 2 2 x x 1 − + O 2 n n = ( ( )) 2 2 x x 1 + + O 2 n n ( ) 2 6 x x = 1 − + O 2 n n ( ( )) 2 6 x x = exp − + O . 2 n n ©2026 HMMT Therefore, we deduce that ( ( )) 2 2 f ( n + x ) 6 1 + · · · + x = exp − (1 + 2 + · · · + x ) + O 2 f ( n ) n n ( ( )) 2 3 3 x x = exp − + O . 2 n n ∑ 3 n/ 2 We now estimate the sum f ( k ) by splitting into two parts. k =0 3 x 0 . 6 • If | k − n | > n , then letting x = k − n , then we have that the error term − is much smaller 2 n ( ) 2 3 x 0 . 2 than the main term − . In particular, we get that f ( n + x ) ≤ exp − n f ( n ) , so we get that n ∑ 0 . 2 f ( k ) ≤ n exp( − n ) f ( n ) ≪ f ( n ) , 0 . 6 | k − n | >n so this part of the sum is dominated by the maximum term f ( n ) . 0 . 6 2 • If | k − n | < n , then if x = k − n , then f ( k ) /f ( n ) = (1 + o (1)) exp( − 3 x /n ) . Therefore, we have that 0 . 6 ( ) n 2 ∑ ∑ 3 x f ( k ) ∼ f ( n ) exp − . n 0 . 6 0 . 6 | k − n | <n x = − n 2 − 3 x /n To estimate this sum, we can bound it by an integral. Since the function g ( x ) = e is monotone on x ≤ 0 and x ≥ 0 , and g (0) = 1 , we get that 0 . 6 ( ) 0 . 6 ( ) ∫ n n 2 2 ∑ 3 x 3 x exp − = exp − dx + O (1) . n n 0 . 6 − n 0 . 6 x = − n Finally, we estimate the integral by extending it to infinity and estimating its tail. We get that 0 . 6 ( ) 0 . 6 ( ) ∫ n n 2 2 ∑ 3 x 3 x exp − = exp − dx + O (1) . n n 0 . 6 − n 0 . 6 x = − n ( ) ( ) ∫ ∫ ∞ ∞ 2 2 3 x 3 x = exp − dx + 2 exp − dx + O (1) n n 0 . 6 −∞ n √ (∫ ( ) ) ∞ 2 πn 3 x = + O x exp − dx + O (1) 3 0 . 6 n n √ ( ( )) 2 · 0 . 6 πn n 3 · n = + O exp − + O (1) 3 6 n √ πn = + O (1) . 3 Hence, we have that (√ ) ∑ πn f ( k ) ∼ + O (1) f ( n ) . 3 0 . 6 | k − n | <n Thus, by combining two cases, we have that √ 3 n/ 2 ∑ πn f ( k ) ∼ f ( n ) . 3 k =0 (3 n )! Finally, since f ( n ) = , we can use Stirling’s formula as above to get the claimed asymptotic 3 n ! formula. ©2026 HMMT Remark. This sum is the number of possible moves that Marin can do in Team Round Problem 8 when n = 30000 .