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

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== Solution ==
 
== Solution ==
  
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We observe that <math>f(1,0) = f(0,1) = 2 </math> and that <math>f(1, y+1) = f(1, f(1,y)) = f(1,y) + 1</math>, so by induction, <math>f(1,y) = y+2 </math>.  Similarly, <math>f(2,0) = f(1,1) = 3</math> and <math>f(2, y+1) = f(2,y) + 2</math>, yielding <math>f(2,y) = 2y + 3</math>.
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We observe that <math>f(1,0) = f(0,1) = 2 </math> and that <math>f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1</math>, so by induction, <math>f(1,y) = y+2 </math>.  Similarly, <math>f(2,0) = f(1,1) = 3</math> and <math>f(2, y+1) = f(2,y) + 2</math>, yielding <math>f(2,y) = 2y + 3</math>.
  
 
We continue with <math>f(3,0) + 3 = 8 </math>; <math>f(3, y+1) + 3 = 2(f(3,y) + 3)</math>; <math>f(3,y) + 3 = 2^{y+3}</math>; and <math>f(4,0) + 3 = 2^{2^2}</math>; <math>f(4,y) + 3 = 2^{f(4,y) + 3}</math>.
 
We continue with <math>f(3,0) + 3 = 8 </math>; <math>f(3, y+1) + 3 = 2(f(3,y) + 3)</math>; <math>f(3,y) + 3 = 2^{y+3}</math>; and <math>f(4,0) + 3 = 2^{2^2}</math>; <math>f(4,y) + 3 = 2^{f(4,y) + 3}</math>.

Revision as of 23:15, 6 January 2023

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
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