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July 26th, 2017, 02:53 PM  #1 
Member Joined: Oct 2015 From: Antarctica Posts: 62 Thanks: 0  For what Value of X will the following conditions be satisfied?
Suppose I have the following six equations: x^60 = y1 x^30 = y2 x^20 = y3 x^15 = y4 x^12 = y5 x^10 = y6 What is the largest value of X (where X is greater than one) such that all y's (y1 through y6) are integers and y1 is smaller than one billion? Obviously the answer is somewhere between one and two... but I don't know if it actually exists (in an exact form). How would you go about solving this problem? Can it even be solved? Last edited by John Travolski; July 26th, 2017 at 02:55 PM. 
July 26th, 2017, 03:09 PM  #2  
Senior Member Joined: May 2016 From: USA Posts: 750 Thanks: 302  Quote:
$1 < x \text { and } x^{60} < 10^9 \implies 1^{60} < x^{60} < 10^9 \implies$ $\log_{10} \left (1^{60} \right ) < \log_{10} \left ( x^{60}\right ) < \log_{10} \left ( 10^9 \right ) \implies WHAT?$ EDIT: I always recommend solving the related equality before solving an inequality. I suspect that if you had set yourself to solve $x^{60} = 10^9$ you might have thought about logarithms on your own. Last edited by skipjack; July 27th, 2017 at 08:41 AM.  
July 26th, 2017, 03:26 PM  #3  
Member Joined: Oct 2015 From: Antarctica Posts: 62 Thanks: 0  Quote:
Since it's (sort of?) a system of equations with a condition revolving around multiple outcomes being integers, I don't see how using logarithms will actually arrive at the solution. Last edited by skipjack; July 27th, 2017 at 08:41 AM.  
July 27th, 2017, 08:40 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 750 Thanks: 302 
Sorry. My first response was careless and not helpful. I have thought about this and still seem unable to help. It is clear that if x exists, then 1 < x < 2. Suppose x is irrational and p is the smallest integer such that $x^p$ is rational. If p is 5, then $x^q$ is rational only if q is a multiple of 5. But that means $x^{12}$ is not rational. If p is 2, then $x^q$ is rational only if q is a multiple of 2. But that means $x^{15}$ is not rational. Thus if x exists, x is rational. And here I come to a halt. 
July 27th, 2017, 08:57 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 17,725 Thanks: 1359 
As x = y4*y6/y5², x is rational.

July 30th, 2017, 09:11 AM  #6 
Member Joined: Oct 2015 From: Antarctica Posts: 62 Thanks: 0 
I'm sorry, but I'm not understanding this. I still don't know if it's possible to arrive at a solution or not.


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