HMMT 十一月 2020 · 冲刺赛 · 第 36 题
HMMT November 2020 — Guts Round — Problem 36
题目详情
- [20] Let p be the i th prime. Let i 50 ∑ i − 1 49 f ( x ) = p x = 2 + 3 x + · · · + 229 x . i i =1 If a is the unique positive real number with f ( a ) = 100, estimate A = b 100000 a c . An estimate of E will earn max(0 , b 20 − | A − E | / 250 c ) points.
解析
- [20] Let p be the i th prime. Let i 50 ∑ i − 1 49 f ( x ) = p x = 2 + 3 x + · · · + 229 x . i i =1 If a is the unique positive real number with f ( a ) = 100, estimate A = b 100000 a c . An estimate of E will earn max(0 , b 20 − | A − E | / 250 c ) points. Proposed by: Daniel Zhu Answer: 83601 Solution: Note f ( x ) is increasing. Since f (0) = 2 and f (1) ≈ 50000, we have 0 < a < 1. Since we know that p = 229, we can crudely bound 50 ∞ ∑ 5 i − 1 f ( x ) . 5 ix = . 2 (1 − x ) i =1 − 1 / 2 Setting this equal to 100 yields x = 1 − 20 ≈ 0 . 78, so this is a good lower bound for a , though just outside the window to receive points. A better estimate can be obtained by noting that since p = 100, it is more accurate to write 25 ∞ ∑ 4 i − 1 f ( x ) . 4 ix = , 2 (1 − x ) i =1 which yields a = 0 . 8, good enough for 5 points. However, we can do better. If we know that a ≈ 0 . 8, the “most significant terms” will occur at the i where p /p ≈ 0 . 8. The first few primes are 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31, so this transition i i +1 ∑ ∞ 19 i − 1 occurs roughly at p = 19. Thus, it is more accurate to approximate f ( x ) = ix , so 8 i =1 8 √ − 1 / 2 a = 1 − 19 / 800 ≈ 1 − 40 ≈ 0 . 85, good enough for 14 points. Repeating this process again with the new estimate for a reveals that p = 23 may have been a better 9 √ √ choice, which yield a = 1 − 23 / 900 ≈ 1 − 0 . 0256 = 0 . 84. This is good enough for 18 points.