Difference between revisions of "1969 Canadian MO Problems/Problem 8"
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== Problem == | == Problem == | ||
| − | Let <math> | + | Let <math>f</math> be a function with the following properties: |
| − | 1) <math> | + | 1) <math>f(n)</math> is defined for every positive integer <math>n</math>; |
| − | 2) <math> | + | 2) <math>f(n)</math> is an integer; |
| − | 3) <math> | + | 3) <math>f(2)=2</math>; |
| − | 4) <math> | + | 4) <math>f(mn)=f(m)f(n)</math> for all <math>m</math> and <math>n</math>; |
| − | 5) <math> | + | 5) <math>f(m)>f(n)</math> whenever <math>m>n</math>. |
| − | Prove that <math> | + | Prove that <math>f(n)=n</math>. |
== Solution == | == Solution == | ||
| − | It's easily shown that <math> | + | It's easily shown that <math>f(1)=1</math> and <math>f(4)=4</math>. Since <math>f(2)<f(3)<f(4),</math> <math>f(3) = 3.</math> |
| − | Now, assume that <math> | + | Now, assume that <math>f(2n+2)=f(2(n+1))=f(2)f(n+1)=2n+2</math> is true for all <math>f(k)</math> where <math>k\leq 2n.</math> |
| − | It follows that <math> | + | It follows that <math>2n<f(2n+1)<2n+2.</math> Hence, <math>f(2n+1)=2n+1</math>, and by induction <math>f(n) = n</math>. |
| − | - | + | {{Old CanadaMO box|num-b=7|num-a=9|year=1969}} |
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Latest revision as of 22:42, 17 November 2007
Problem
Let 
be a function with the following properties:
1) 
is defined for every positive integer 
;
2) is an integer;
3) 
;
4) 
for all 
and ;
5) 
whenever 
.
Prove that 
.
Solution
It's easily shown that 
and 
. Since 
Now, assume that 
is true for all 
where 
It follows that 
Hence, 
, and by induction 
.
| 1969 Canadian MO (Problems) | ||
| Preceded by Problem 7 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 9 |
