Daily Math: Iran Mo 2023
Introduction
Here is my approach to problem 5 in Iran MO 2023 (3rd Round.) Always lots of fun!
This is not a perfect solutions; it is my approach (brief solution), so there might be some logical loophole. Please let me know via comments :)
Check out the problems at here!
Problem 2 (≤15’)
Within seconds of reading the question, I became sure that the answer was NO. After a few tries, I figured out \(g(1)=f(1)=1\), and I could make a key proposition
\(g(n)\) must be \(g(n) = n\), which gives \(f(n)=2n-1\), a contradiction.
I was stucked on trying to prove this proposition for a while. After about ten minutes, I realized I was hanging around to prove even though I already had a key idea. Yuck!
Proof. \(g(1)=1\) (base condition). Assume that for \(1\leq k \leq n\) \(g(k)=k\). Then from \(ng(n)=(n-1)g(n-1)+f(n)\), \(g(n)>\frac{(n-1)^2}{n}\). Because \(g\) is bijective, \(g(n)=n\).
Thus, the proposition is correct and the answer is NO.