Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2012 USAMO Problems/Problem 4"

m (→‎See also)
(→‎See also)
Line 61: Line 61:
  
 
{{USAMO newbox|year=2012|num-b=3|num-a=5}}
 
{{USAMO newbox|year=2012|num-b=3|num-a=5}}
 +
{{MAA Notice}}
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 17:18, 3 July 2013

Problem

Find all functions $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ (where $\mathbb{Z}^+$ is the set of positive integers) such that $f(n!) = f(n)!$ for all positive integers $n$ and such that $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$.

Solution

By the first condition we have $f(1)=f(1!)=f(1)!$ and $f(2)=f(2!)=f(2)!$, so $f(1)=1$ or $2$ and similarly for $f(2)$. By the second condition, we have \[n\cdot n!=(n+1)!-n! \mid f(n+1)!-f(n)! \qquad \qquad (1)\] for all positive integers $n$.

Suppose that for some $n \geq 2$ we have $f(n) = 1$. We claim that $f(k)=1$ for all $k\ge n$. Indeed, from Equation (1) we have $f(n+1)!\equiv 1 \mod n\cdot n!$, and this is only possible if $f(n+1)=1$; the claim follows by induction.

We now divide into cases:

Case 1: $f(1)=f(2)=1$

This gives $f(n)=1$ always from the previous claim, which is a solution.

Case 2: $f(1)=2, f(2)=1$

This implies $f(n)=1$ for all $n\ge 2$, but this does not satisfy the initial conditions. Indeed, we would have \[3-1 \mid f(3)-f(1)\] and so $2\mid -1$, a contradiction.

Case 3: $f(1)=1$, $f(2)=2$

We claim $f(n)=n$ always by induction. The base cases are $n = 1$ and $n = 2$. Fix $k > 1$ and suppose that $f(k)=k$. By Equation (1) we have that \[f(k+1)! \equiv k! \mod k\cdot k! .\] This implies $f(k+1)<2k$ (otherwise $f(k+1)!\equiv 0 \mod k\cdot k!$). Also we have \[(k+1)-1 \mid f(k+1)-f(1)\] so $f(k+1)\equiv 1 \mod k$. This gives the solutions $f(k+1)=1$ and $f(k+1)=k+1$. The first case is obviously impossible, so $f(k + 1) = k + 1$, as desired. By induction, $f(n) = n$ for all $n$. This also satisfies the requirements.

Case 4: $f(1)=f(2)=2$

We claim $f(n)=2$ by a similar induction. Again if $f(k)=2$, then by (1) we have \[f(k+1)\equiv 2 \mod k\cdot k!\] and so $f(k+1)<2k$. Also note that \[k+1-1 \mid f(k+1)-2\] and \[k+1-2 \mid f(k+1)-2\] so $f(k+1)\equiv 2 \mod k(k-1)$. Then the only possible solution is $f(k+1)=2$. By induction, $f(n) = 2$ for all $n$, and this satisfies all requirements.


In summary, there are three solutions: $\boxed{f(n)=1, f(n)=2, f(n)=n}$.

See also

2012 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_USAMO_Problems/Problem_4&oldid=53399"