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

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

Revision as of 12:53, 27 March 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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