1988 AIME Problems/Problem 2

Revision as of 01:19, 6 June 2014 by Mathcool2009 (talk | contribs) (Solution)

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 $f_{1}(11)=4$

$f_2(11) = f_1(4)=16$

$f_3(11) = f_1(16)=49$

$f_4(11) = f_1(49)=169$

$f_5(11) = f_1(169)=256$

$f_6(11) = f_1(256)=169$

Note that this revolves between the two numbers. $f_{1988}(169)=169\implies f_{1988}(11)=\boxed{169}$

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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