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 July 26th, 2017, 02:53 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 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
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 Originally Posted by John Travolski 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? How would you go about solving this problem? Can it even be solved?
It is obvious that x = 0 will work. It is also obvious that 0 < x < 1 will not work. And finally it is obvious that x = 1 will work. Can it work if x > 1.

$1 < x \text { and } x^{60} < 10^9 \implies 1^{60} < x^{60} < 10^9 \implies$

$\log_{10} \left (1^{60} \right ) < \log_{10} \left ( x^{60}\right ) < \log_{10} \left ( 10^9 \right ) \implies WHAT?$

EDIT: I always recommend solving the related equality before solving an inequality. I suspect that if you had set yourself to solve

$x^{60} = 10^9$

Last edited by skipjack; July 27th, 2017 at 08:41 AM.

July 26th, 2017, 03:26 PM   #3
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 Originally Posted by JeffM1 It is obvious that x = 0 will work. It is also obvious that 0 < x < 1 will not work. And finally it is obvious that x = 1 will work. Can it work if x > 1. $1 < x \text { and } x^{60} < 10^9 \implies 1^{60} < x^{60} < 10^9 \implies$ $\log_{10} \left (1^{60} \right ) < \log_{10} \left ( x^{60}\right ) < \log_{10} \left ( 10^9 \right ) \implies WHAT?$ 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.
Yes, I know that I can use logarithms to solve any individual one of those equations where Yn is an integer, but the value of X must be the same value for all six equations. Furthermore. Keeping y1 through y6 as an integer while y1 is less than a billion is what's difficult since X needs to be greater than one.

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: 1,310 Thanks: 552 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: 21,028 Thanks: 2259 As x = y4*y6/y5², x is rational. Thanks from JeffM1
 July 30th, 2017, 09:11 AM #6 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 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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