AIME 1988 · 第 2 题 AIME 1988 — Problem 2 专题 Contest Math 难度 L4 来源 AIME 题目详情 Problem For any positive integer kkk, let f1(k)f_1(k)f1(k) denote the square of the sum of the digits of kkk. For n≥2n \ge 2n≥2, let fn(k)=f1(fn−1(k))f_n(k) = f_1(f_{n - 1}(k))fn(k)=f1(fn−1(k)). Find f1988(11)f_{1988}(11)f1988(11). 解析 Solution We see that f1(11)=4f_{1}(11)=4f1(11)=4 f2(11)=f1(4)=16f_2(11) = f_1(4)=16f2(11)=f1(4)=16 f3(11)=f1(16)=49f_3(11) = f_1(16)=49f3(11)=f1(16)=49 f4(11)=f1(49)=169f_4(11) = f_1(49)=169f4(11)=f1(49)=169 f5(11)=f1(169)=256f_5(11) = f_1(169)=256f5(11)=f1(169)=256 f6(11)=f1(256)=169f_6(11) = f_1(256)=169f6(11)=f1(256)=169 Note that this revolves between the two numbers. Since 198819881988 is even, we thus have f1988(11)=f4(11)=169f_{1988}(11) = f_{4}(11) = \boxed{169}f1988(11)=f4(11)=169.