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

(Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== 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.
 
Note that this revolves between the two numbers.

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

See also

1988 AIME (ProblemsAnswer KeyResources)
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

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