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

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== Solution ==
 
== Solution ==
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We see that <math>f(11)=4</math>
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We see that <math>f_{1}(11)=4</math>
  
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<math>f(4)=16</math>
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<math>f_2(11) = f_1(4)=16</math>
  
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<math>f(16)=49</math>
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<math>f_3(11) = f_1(16)=49</math>
  
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<math>f(49)=169</math>
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<math>f_4(11) = f_1(49)=169</math>
  
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<math>f(169)=256</math>
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<math>f_5(11) = f_1(169)=256</math>
  
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<math>f(256)=169</math>
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<math>f_6(11) = f_1(256)=169</math>
  
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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>.
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<math>f_{1988}(169)=\boxed{169}</math>
 
  
 
== See also ==
 
== See also ==
 
{{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 (Problems • Answer Key • Resources)
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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