HMMT 二月 2022 · 冲刺赛 · 第 4 题
HMMT February 2022 — Guts Round — Problem 4
题目详情
- [5] For a real number x , let [ x ] be x rounded to the nearest integer and ⟨ x ⟩ be x rounded to the nearest tenth. Real numbers a and b satisfy ⟨ a ⟩ + [ b ] = 98 . 6 and [ a ] + ⟨ b ⟩ = 99 . 3. Compute the minimum possible value of [10( a + b )]. (Here, any number equally between two integers or tenths of integers, respectively, is rounded up. For example, [ − 4 . 5] = − 4 and ⟨ 4 . 35 ⟩ = 4 . 4.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2022, February 19, 2022 — GUTS ROUND Organization Team Team ID#
解析
4 . 39 · 10 . However, since 858522 is relatively close to 796921, the first case is unrealistic, meaning that the second case must be true. 12 Now, looking mod 2, we find that f (796921) ≈ 1 . 32 · 10 . Additionally, we find that mod 5, f (1) ≡ f (3) ≡ 0 (mod 5), so f ( x ) ≡ a ( x − 1)( x − 3) (mod 5). Modulo 5, we now have { 3 a, 4 a } = { f (0) , f (2) } = 12 12 { 2 , 4 } , so it follows that a ≡ 3 (mod 5), f (349710) ≈ 1 . 78 · 10 and f (858522) ≈ 1 . 98 · 10 . There are several ways to finish from here. One (somewhat tedious) method is to use mod 9, which tells us that f (7) = 7, f (5) = 8, f (3) = 4, which tells you that a ≡ 5 (mod 9) (take a finite difference). This tells you that a ≡ 23 (mod 45), and a ≥ 68 can be ruled out for being too large. Another method is to work with the numbers themselves. One way to do this is to note that for quadratic polynomials, x + y f ( y ) − f ( x ) ′ f = . 2 y − x ′ Using this for { 177883 , 348710 } and { 796921 , 858522 } , we find that f (260000) ≈ − 1500000 and ′ ′ f (830000) ≈ 1000000. Thus f (which we know must be linear with slope 2 a ) has slope just less than 50. 2 Either way, we find that a = 23. The actual polynomial is 8529708870514 − 27370172 x + 23 x .