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

Revision as of 18:10, 4 July 2013

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. $f_{1988}(169)=169\implies f_{1988}(11)=\boxed{169}$

@@ Line 20: / Line 20: @@
 == See also ==
 {{AIME box|year=1988|num-b=1|num-a=3}}
+{{MAA Notice}}

1988 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1988 AIME Problems/Problem 2"

Revision as of 18:10, 4 July 2013

Problem

Solution

See also