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

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== See also ==
 
== See also ==
* [[1988 AIME Problems]]
 
 
 
{{AIME box|year=1988|num-b=1|num-a=3}}
 
{{AIME box|year=1988|num-b=1|num-a=3}}

Revision as of 11:22, 29 September 2007

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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