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> | ||
− | 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(3,0)=5, f(3,1)= | + | 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(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> | + | </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)=</cmath> |
Revision as of 21:13, 11 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.
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 |
https://beastacademy.com/ https://beastacademy.com/
Solution 2
We can start by creating a list consisting of certain x an y values and their outputs.
This pattern can be proved using induction. After proving, we continue to setting a list when . This pattern can also be proved using induction. The pattern seems consistent in a common difference of 1. Moving on to All of the numbers are being expressed in the form of where . Lastly where x=4 we have