Difference between revisions of "1981 IMO Problems/Problem 6"
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It follows that <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> when there are 1984 2s, Q.E.D. | It follows that <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> when there are 1984 2s, Q.E.D. | ||
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== Solution 2 == | == Solution 2 == | ||
| − | + | 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>. <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 d up of a common difference of 1. Moving on to <math>x=3</math> <cmath>f(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125</cmath> All of the numbers are being expressed in the form of <math>3^a -3 | ||
| + | </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 | ||
| + | {{alternate solutions}} | ||
| − | + | {{IMO box|num-b=5|after=Last question|year=1981}} | |
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Latest revision as of 17:15, 12 April 2024
Problem
The function 
satisfies
(1) 
(2) 
(3) 
for all non-negative integers 
. Determine 
.
Solution
We observe that 
and that 
, so by induction, 
. Similarly, 
and 
, yielding 
.
We continue with 
; 
; 
; and 
; 
.
It follows that 
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. ![\[f(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5\]](http://latex.artofproblemsolving.com/9/2/e/92ee4f88458bfa8e3d7236c5b9f666bc9fb50b21.png)
This pattern can be proved using induction. After proving, we continue to setting a list when 
. ![\[f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6\]](http://latex.artofproblemsolving.com/6/5/1/65130767fa649bd07215e0316ab1986aa6544e31.png)
This pattern can also be proved using induction. The pattern seems d up of a common difference of 1. Moving on to 
![\[f(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125\]](http://latex.artofproblemsolving.com/3/9/9/399bfbd77170f86c9a3d8b68ca3d2a74c05e49cb.png)
All of the numbers are being expressed in the form of 
where 
. Lastly where x=4 we have ![\[f(4,0)=13, f(4,1)=65533, f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2\]](http://latex.artofproblemsolving.com/2/8/d/28dae2f5ccd6cd71c3da6a67bda9679f4c469036.png)
where each term can be represented as 
when 
. In represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where 
amount of 2s. So therefore the answer is with 1984 2s, 2 tetration 1983.
-Multpi12
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 | ||
