1988 AIME Problems/Problem 2

Revision as of 12:53, 27 March 2007 by Diophantient (talk | contribs) (Problem)

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. $f_{1984}(169)=169$

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions
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