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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 kk, let f1(k)f_1(k) denote the square of the sum of the digits of kk. For n2n \ge 2, let fn(k)=f1(fn1(k))f_n(k) = f_1(f_{n - 1}(k)). Find f1988(11)f_{1988}(11).

解析

Solution

We see that f1(11)=4f_{1}(11)=4

f2(11)=f1(4)=16f_2(11) = f_1(4)=16 f3(11)=f1(16)=49f_3(11) = f_1(16)=49 f4(11)=f1(49)=169f_4(11) = f_1(49)=169 f5(11)=f1(169)=256f_5(11) = f_1(169)=256 f6(11)=f1(256)=169f_6(11) = f_1(256)=169

Note that this revolves between the two numbers. Since 19881988 is even, we thus have f1988(11)=f4(11)=169f_{1988}(11) = f_{4}(11) = \boxed{169}.