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

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</math> where <math>a=y+3</math>. Lastly where x=4 we have <cmath>f(4,0)=13,  f(4,1)=65533,  f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2</cmath> where each term can be represented as <math>^a 2</math> when <math>a=y+2</math>. In <math>^a 2</math> represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where <math>a</math> amount of 2s. So therefore the answer is <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> with 1984 2s, 2  tetration 1983.
 
</math> where <math>a=y+3</math>. Lastly where x=4 we have <cmath>f(4,0)=13,  f(4,1)=65533,  f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2</cmath> where each term can be represented as <math>^a 2</math> when <math>a=y+2</math>. In <math>^a 2</math> represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where <math>a</math> amount of 2s. So therefore the answer is <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> with 1984 2s, 2  tetration 1983.
 
-Multpi12
 
-Multpi12
 +
 +
 +
<b>Remark.</b> This function is well-studied, known widely as the Ackermann function.
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
 
{{IMO box|num-b=5|after=Last question|year=1981}}
 
{{IMO box|num-b=5|after=Last question|year=1981}}

Latest revision as of 02:45, 5 July 2025

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 $f(4,0) + 3 = 2^{2^2}$; $f(4,y) + 3 = 2^{f(4,y-1) + 3}$.

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

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 d up of a common difference of 1. Moving on to $x=3$ \[f(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125\] All of the numbers are being expressed in the form of $3^a -3$ where $a=y+3$. Lastly where x=4 we have \[f(4,0)=13, f(4,1)=65533, f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2\] where each term can be represented as $^a 2$ when $a=y+2$. In $^a 2$ represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where $a$ amount of 2s. So therefore the answer is $f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3$ with 1984 2s, 2 tetration 1983. -Multpi12


Remark. This function is well-studied, known widely as the Ackermann function.

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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