Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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 (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_2&oldid=14221"