1997 AIME 第 9 题

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9.

给定一个非负实数 xx,令 x\langle x\rangle 表示 xx 的小数部分;也就是说, x=xx\langle x\rangle = x - \lfloor x\rfloor,其中 x\lfloor x\rfloor 表示不超过 xx 的最大整数。设 aa 为正数,a1=a2\langle a^{-1}\rangle = \langle a^2\rangle,且 2<a2<32 \lt a^2 \lt 3。求 a12144a1a^{12} - 144a^{-1} 的值。

Given a nonnegative real number x,x, let x\langle x\rangle denote the fractional part of x;x; that is, x=xx,\langle x\rangle = x - \lfloor x\rfloor, where x\lfloor x\rfloor denotes the greatest integer less than or equal to x.x. Suppose that aa is positive, a1=a2,\langle a^{-1}\rangle = \langle a^2\rangle, and 2<a2<3.2 \lt a^2 \lt 3. Find the value of a12144a1.a^{12} - 144a^{-1}.

答案:233
知识点:取整函数因式分解代数变形
难度评级:2560
解答:

2<a2<32 \lt a^2 \lt 32<a<3\sqrt{2} \lt a \lt \sqrt{3},所以 0<a1<10 \lt a^{-1} \lt 1,从而 a1=a1\langle a^{-1}\rangle = a^{-1},同时 a2=a22\langle a^2\rangle = a^2 - 2。条件变为 a1=a22a^{-1} = a^2 - 2,即 a32a1=0a^3 - 2a - 1 = 0,它可分解为 因为 a>0a \gt 0,所以 a=1+52a = \frac{1 + \sqrt{5}}{2},也就是黄金比例;确实有 a2=a+12.618a^2 = a + 1 \approx 2.618,位于 (2,3)(2, 3) 内。 (a+1)(a2a1)=0.(a + 1)(a^2 - a - 1) = 0.

反复使用 a2=a+1a^2 = a + 1a4=(a+1)2=3a+2a^4 = (a+1)^2 = 3a + 2a8=(3a+2)2a^8 = (3a+2)^2 =9(a+1)+12a+4= 9(a+1) + 12a + 4 =21a+13= 21a + 13,且 a12=a8a4a^{12} = a^8 a^4 =(21a+13)(3a+2)= (21a + 13)(3a + 2) =63(a+1)+81a+26= 63(a+1) + 81a + 26 =144a+89= 144a + 89。又由 a2=a+1a^2 = a + 1a1=a1a^{-1} = a - 1

因此 a12144a1a^{12} - 144a^{-1} =144a+89144(a1)= 144a + 89 - 144(a - 1) =89+144=233= 89 + 144 = 233

From 2<a2<32 \lt a^2 \lt 3 we get 2<a<3,\sqrt{2} \lt a \lt \sqrt{3}, so 0<a1<10 \lt a^{-1} \lt 1 and a1=a1,\langle a^{-1}\rangle = a^{-1}, while a2=a22.\langle a^2\rangle = a^2 - 2. The condition becomes a1=a22,a^{-1} = a^2 - 2, i.e. a32a1=0,a^3 - 2a - 1 = 0, which factors as (a+1)(a2a1)=0.(a + 1)(a^2 - a - 1) = 0. Since a>0,a \gt 0, we get a=1+52,a = \frac{1 + \sqrt{5}}{2}, the golden ratio, and indeed a2=a+12.618a^2 = a + 1 \approx 2.618 lies in (2,3).(2, 3).

Using a2=a+1a^2 = a + 1 repeatedly: a4=(a+1)2=3a+2,a^4 = (a+1)^2 = 3a + 2, a8=(3a+2)2a^8 = (3a+2)^2 =9(a+1)+12a+4= 9(a+1) + 12a + 4 =21a+13,= 21a + 13, and a12=a8a4a^{12} = a^8 a^4 =(21a+13)(3a+2)= (21a + 13)(3a + 2) =63(a+1)+81a+26= 63(a+1) + 81a + 26 =144a+89.= 144a + 89. Also a1=a1a^{-1} = a - 1 from a2=a+1.a^2 = a + 1.

Therefore a12144a1a^{12} - 144a^{-1} =144a+89144(a1)= 144a + 89 - 144(a - 1) =89+144=233.= 89 + 144 = 233.

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