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

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

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 == Problem ==
+For any positive integer <math>k</math>, let <math>f_1(k)</math> denote the square of the sum of the digits of <math>k</math>.  For <math>n \ge 2</math>, let <math>f_n(k) = f_1(f_{n - 1}(k))</math>.  Find <math>f_{1988}(11)</math>.
 == Solution ==

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

Revision as of 12:53, 27 March 2007

Problem

Solution

See also