All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Interesting Function (2) (Posted on 2004-09-03) Difficulty: 4 of 5
Given: f is a function with domain and range of the positive integers, and f satisfies these two conditions:

(1) f(n+1) > f(n); that is, f is strictly increasing, and

(2) f(f(n)) = 3n

Find all possible values of f(955)

See The Solution Submitted by SilverKnight    
Rating: 3.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Steps... | Comment 1 of 6
If f(1)=1, then f(f(1))=3, but then f(1)=3.

If f(1)=3, then f(f(1))=3, but then f(3)=3, and the function wouldn't be strictly increasing.

If f(1)=3+K, then f(f(1))=f(3+K)=3, but f would be decreasing.

So, the only possibility is f(1)=2. Thus, f(f(1))=f(2)=3, f(f(2))=f(3)=6, f(6)=9, f(9)=18, f(18)=27, and so on.

THere are no chances but f(4)=7 and f(5)=8, and then f(7)=12, f(8)=15, f(12)=21, f(15)=24. It seems that f(10)=19 and f(11)=20, to fill out spaces.

After many such experiments, it seems that f(x)=x+1 for x=1 to 2; x+3 for x=3 to 6; x+9 for x=9 to 18; x+27 for x=27 to 54, and so on. Thus f(x)=x+729 for x=729 to 1458, and f(955)=1684.
  Posted by Federico Kereki on 2004-09-03 13:12:18

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information