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

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We can start by creating a list consisting of certain x an y values and their outputs. <cmath>f(0,0)=1,  f(0,1)=2,  f(0,2)=3, f(0,3)=4, f(0,4)=5</cmath>
 
We can start by creating a list consisting of certain x an y values and their outputs. <cmath>f(0,0)=1,  f(0,1)=2,  f(0,2)=3, f(0,3)=4, f(0,4)=5</cmath>
  
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This pattern can be proved using induction. After proving, we continue to setting a list when <math>x=2</math> <math><cmath>cat</cmath></math>
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This pattern can be proved using induction. After proving, we continue to setting a list when <math>x=2</math>. <cmath>f(1,0)=2,  f(1,1)=3,  f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath> This pattern can also be proved using induction. The pattern seems consistent in a common difference of 1. Moving on to <math>x=3</math> <cmath>f(1,0)=2,  f(1,1)=3,  f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath>

Revision as of 21:06, 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
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Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
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All IMO Problems and Solutions

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Solution 2

We can start by creating a list consisting of certain x an y values and their outputs. \[f(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5\]

This pattern can be proved using induction. After proving, we continue to setting a list when $x=2$. \[f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6\] This pattern can also be proved using induction. The pattern seems consistent in a common difference of 1. Moving on to $x=3$ \[f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6\]

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