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Difference between revisions of "1981 IMO Problems/Problem 6"

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{{IMO box|num-b=5|after=Last question|year=1981}}
 
{{IMO box|num-b=5|after=Last question|year=1981}}
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== Solution 2 ==
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== Solution 2 ==https://beastacademy.com/
  
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\documentclass{article}
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\begin{center}
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\usepackage{multirow}
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\begin{tabular}{||c c c c||}  
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\begin{document}
 
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\begin{tabular}{ |p{3cm}||p{3cm}|p{3cm}|p{3cm}| }
 
 
  \hline
 
  \hline
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  \multicolumn{4}{|c|}{Country List} \\
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  Col1 & Col2 & Col2 & Col3 \\ [0.5ex]
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\hline\hline
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1 & 6 & 87837 & 787 \\  
 
  \hline
 
  \hline
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  Country Name or Area Name& ISO ALPHA 2 Code &ISO ALPHA 3 Code&ISO numeric Code\\
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  2 & 7 & 78 & 5415 \\
 
  \hline
 
  \hline
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  Afghanistan  & AF    &AFG&   004\\
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  3 & 545 & 778 & 7507 \\
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  Aland Islands&  AX  & ALA  &248\\
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  \hline
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Albania &AL & ALB&  008\\
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  4 & 545 & 18744 & 7560 \\
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Algeria    &DZ & DZA&  012\\
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  \hline
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  American Samoa&   AS  & ASM&016\\
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  5 & 88 & 788 & 6344 \\ [1ex]
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  Andorra& AD  & AND  &020\\
 
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  Angola& AO  & AGO&024\\
 
 
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\end{tabular}
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\end{document}
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Revision as of 20:40, 8 April 2024

Problem

The function $f(x,y)$ satisfies

(1) $f(0,y)=y+1,$

(2) $f(x+1,0)=f(x,1),$

(3) $f(x+1,y+1)=f(x,f(x+1,y)),$

for all non-negative integers $x,y$. Determine $f(4,1981)$.

Solution

We observe that $f(1,0) = f(0,1) = 2$ and that $f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1$, so by induction, $f(1,y) = y+2$. Similarly, $f(2,0) = f(1,1) = 3$ and $f(2, y+1) = f(2,y) + 2$, yielding $f(2,y) = 2y + 3$.

We continue with $f(3,0) + 3 = 8$; $f(3, y+1) + 3 = 2(f(3,y) + 3)$; $f(3,y) + 3 = 2^{y+3}$; and $f(4,0) + 3 = 2^{2^2}$; $f(4,y) + 3 = 2^{f(4,y) + 3}$.

It follows that $f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3$ when there are 1984 2s, Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1981 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last question
All IMO Problems and Solutions

== Solution 2 ==https://beastacademy.com/

\begin{center} \begin{tabular}{||c c c c||} 

\hline
Col1 & Col2 & Col2 & Col3 \\ [0.5ex] 
\hline\hline
1 & 6 & 87837 & 787 \\ 
\hline
2 & 7 & 78 & 5415 \\
\hline
3 & 545 & 778 & 7507 \\
\hline
4 & 545 & 18744 & 7560 \\
\hline
5 & 88 & 788 & 6344 \\ [1ex] 
\hline

\end{tabular} \end{center}

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