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"

m (→‎See also)
m (→‎Solution)
Line 16: Line 16:
  
 
Note that this revolves between the two numbers.
 
Note that this revolves between the two numbers.
−
<math>f_{1984}(169)=169</math>
+
<math>f_{1988}(169)=\boxed{169}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1988|num-b=1|num-a=3}}
 
{{AIME box|year=1988|num-b=1|num-a=3}}

Revision as of 22:00, 26 May 2009

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. $f_{1988}(169)=\boxed{169}$

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=31883"