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

(Problem)
(Solution)
Line 4: Line 4:
 
== Solution ==
 
== Solution ==
 
We see that <math>f(11)=4</math>
 
We see that <math>f(11)=4</math>
 +
 
<math>f(4)=16</math>
 
<math>f(4)=16</math>
 +
 
<math>f(16)=49</math>
 
<math>f(16)=49</math>
 +
 
<math>f(49)=169</math>
 
<math>f(49)=169</math>
 +
 
<math>f(169)=256</math>
 
<math>f(169)=256</math>
 +
 
<math>f(256)=169</math>
 
<math>f(256)=169</math>
 +
 
Note that this revolves between the two numbers.
 
Note that this revolves between the two numbers.
 
<math>f_{1984}(169)=169</math>
 
<math>f_{1984}(169)=169</math>

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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