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

 
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It follows that <math>\displaystyle f(4,1981) = 2^{2\cdot ^{ . \cdot 2}}</math> when there are 1984 2s, Q.E.D.
 
It follows that <math>\displaystyle f(4,1981) = 2^{2\cdot ^{ . \cdot 2}}</math> when there are 1984 2s, Q.E.D.
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{{alternate solutions}}
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== Resources ==
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* [[1981 IMO Problems]]
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366648#p366648 AoPS/MathLinks Discussion]
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[[Category:Olympiad Algebra Problems]]

Revision as of 17:36, 29 October 2006

Problem

The function $\displaystyle f(x,y)$ satisfies

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

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

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

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

Solution

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

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

It follows that $\displaystyle f(4,1981) = 2^{2\cdot ^{ . \cdot 2}}$ 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.

Resources

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