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

Revision as of 01:19, 6 June 2014

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 $f_{1}(11)=4$

$f_2(11) = f_1(4)=16$

$f_3(11) = f_1(16)=49$

$f_4(11) = f_1(49)=169$

$f_5(11) = f_1(169)=256$

$f_6(11) = f_1(256)=169$

Note that this revolves between the two numbers. $f_{1988}(169)=169\implies f_{1988}(11)=\boxed{169}$

@@ Line 3: / Line 3: @@
 == Solution ==
-We see that <math>f(11)=4</math>
+We see that <math>f_{1}(11)=4</math>
-<math>f(4)=16</math>
+<math>f_2(11) = f_1(4)=16</math>
-<math>f(16)=49</math>
+<math>f_3(11) = f_1(16)=49</math>
-<math>f(49)=169</math>
+<math>f_4(11) = f_1(49)=169</math>
-<math>f(169)=256</math>
+<math>f_5(11) = f_1(169)=256</math>
-<math>f(256)=169</math>
+<math>f_6(11) = f_1(256)=169</math>
 Note that this revolves between the two numbers.

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

Revision as of 01:19, 6 June 2014

Problem

Solution

See also