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

Home > Just Math
Exponent Crossed Power (Posted on 2009-03-18) Difficulty: 3 of 5
Determine all possible pair(s) (X, Y) of positive integers that satisfy this equation.

                                        XXX = YY

Note: The order of calculation in XXX is as given in this article.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts going farther | Comment 5 of 10 |
(In reply to improved program by Charlie)

Rather than incrementing y by 1 to get to ever higher values needed for y, it's better to calculate the approximate needed y for a given x, and try the integer below and the integer above. That way, the following program goes up to x=800, needing values of y with over 2000 digits:

 10   point 16
 20   for X=2 to 800
 30    Lhs=(X^X)*log(X)
 40    Ly=100:PrevY=2:Y=2
 50    loop
 60      Y=abs(Lhs/log(Y))
 70      if abs(Y-PrevY)<0.0000000000000000000001 then goto *FoundOne
 80      PrevY=abs(Y)
 90    endloop
100   *FoundOne
110    Y1=int(Y):Y2=-int(-Y)
115    if abs(Lhs-Y1*log(Y1))<0.000000001 then print X,Y,Lhs,Y1*log(Y1)
116    if abs(Lhs-Y2*log(Y2))<0.000000001 then print X,Y,Lhs,Y2*log(Y2)
200   next

It still does not find any pairs beyond (1,1).


  Posted by Charlie on 2009-03-18 15:50:28
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information