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

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== Solution ==
 
== Solution ==
We see that <math>f(11)=4</math>
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We see that <math>f_{1}(11)=4</math>
  
<math>f(4)=16</math>
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<math>f_2(11) = f_1(4)=16</math>
  
<math>f(16)=49</math>
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<math>f_3(11) = f_1(16)=49</math>
  
<math>f(49)=169</math>
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<math>f_4(11) = f_1(49)=169</math>
  
<math>f(169)=256</math>
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<math>f_5(11) = f_1(169)=256</math>
  
<math>f(256)=169</math>
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<math>f_6(11) = f_1(256)=169</math>
  
Note that this revolves between the two numbers.
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Note that this revolves between the two numbers. Since <math>1988</math> is even, we thus have <math>f_{1988}(11) = f_{4}(11) = \boxed{169}</math>.
<math>f_{1984}(169)=169</math>
 
  
 
== See also ==
 
== See also ==
* [[1988 AIME Problems]]
 
 
 
{{AIME box|year=1988|num-b=1|num-a=3}}
 
{{AIME box|year=1988|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 18:16, 27 February 2018

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. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$.

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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