1982 IMO Problems/Problem 1
Problem
The function is defined on the positive integers and takes non-negative integer values. and for all Determine .
Solution 1
Clearly so .Contradiction!So .This forces .Hence so the inequality forces .Now (Note:This is valid for or ).Contradiction!Hence the non-decreasing nature of gives .Hence .
So .
This solution was posted and copyrighted by sayantanchakraborty. The original thread for this problem can be found here: [1]
Solution 2
First observe that
Since is a positive integer, we need . Next, observe that
This solution was posted and copyrighted by Solumilkyu. The original thread for this problem can be found here: [2]
Solution 3
We show that for , where [ ] denotes the integral part. We show first that . must be , otherwise would be negative. Hence or = or . But we are told , so . It follows by induction that . For + or or . Moreover if we ever get , then the same argument shows that for all . But , so for all . Now or = or . But , so . Hence Similarly, . In particular .
This solution was posted and copyrighted by Tega. The original thread for this problem can be found here: [3]
Solution 4
Similar to solution 3.
Proof: Lemma 1: Let, be assertion. Similarly,we can induct to get . Lemma proved.
Then we see that, Then, Then we can easily get,by assertion Hence, .And, we are done.
This solution was posted and copyrighted by IMO2019. The original thread for this problem can be found here: [4]
See Also
1982 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |