1988 AIME Problems/Problem 2: Difference between revisions

Revision as of 12:52, 27 March 2007

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. $f_{1984}(169)=169$

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