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1981 IMO Problems/Problem 6: Difference between revisions

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{{IMO box|num-b=5|after=Last question|year=1981}}
{{IMO box|num-b=5|after=Last question|year=1981}}
== Solution 2 ==
== Solution 2 ==
\documentclass{article}
\usepackage{multirow}
\begin{document}
\begin{tabular}{ |p{3cm}||p{3cm}|p{3cm}|p{3cm}|  }
\hline
\multicolumn{4}{|c|}{f(x,y)} \\
\hline
x value & y=0 & y=1 & y=2 \\
\hline
    & AF    &AFG&  004\\
Aland Islands&  AX  & ALA  &248\\
Albania &AL & ALB&  008\\
Algeria    &DZ & DZA&  012\\
American Samoa&  AS  & ASM&016\\
Andorra& AD  & AND  &020\\
Angola& AO  & AGO&024\\
\hline
\end{tabular}
\end{document}




Revision as of 20:38, 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

\documentclass{article} \usepackage{multirow} \begin{document} \begin{tabular}{ |p{3cm}||p{3cm}|p{3cm}|p{3cm}| } 

\hline
\multicolumn{4}{|c|}{f(x,y)} \\
\hline
x value & y=0 & y=1 & y=2 \\
\hline
   & AF    &AFG&   004\\
Aland Islands&   AX  & ALA   &248\\
Albania &AL & ALB&  008\\
Algeria    &DZ & DZA&  012\\
American Samoa&   AS  & ASM&016\\
Andorra& AD  & AND   &020\\
Angola& AO  & AGO&024\\
\hline

\end{tabular} \end{document}

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