1981 IMO Problems/Problem 6
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.
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 d up of 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 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.
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