# Difference between revisions of "1988 AIME Problems/Problem 2"

## Problem

For any positive integer $k$, let $f_1(k)$ denote the square of the sum of the digits of $k$. For $n \ge 2$, let $f_n(k) = f_1(f_{n - 1}(k))$. Find $f_{1988}(11)$.

## Solution

We see that $f(11)=4$ $f(4)=16$ $f(16)=49$ $f(49)=169$ $f(169)=256$ $f(256)=169$ Note that this revolves between the two numbers. $f_{1984}(169)=169$