Competition problem — Number theory

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Number theory Difficulty 2.9 Junior Find the answer

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 $f_{1988}(11)$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.