Solving for 100x: A Math Puzzle

  • Thread starter Thread starter thomas49th
  • Start date Start date
  • Tags Tags
    Puzzle
AI Thread Summary
The discussion centers on converting repeating, non-terminating decimals into fractional form using the equation x = 0.91919191. The method involves multiplying by 100 to isolate the repeating part, leading to the equation 100x - x = 91, which simplifies to x = 91/99. This technique serves as a constructive proof that periodic decimals are rational numbers, demonstrating that they can always be expressed as fractions between two natural numbers. Understanding this method is essential for rewriting periodic decimals in a more manageable format. The focus is on the mathematical principle rather than practical applications.
thomas49th
Messages
645
Reaction score
0
Hi, I recall a thingy when you have somthing like

x = 0.91919191

then you go 100x = 91.919191
100x - x = 91

When do you use this? It's in conjuction with somthing else... looked in my textbook can't find it.

Thanks
 
Physics news on Phys.org
Personally, I use that when I have a repeating, non-terminating decimal that I want to get in fractional form so it looks prettier in an equation.
 
Why should you care "when is it used"?
It is a wrong question; the technique is never "used"; rather, it illustrates how you may rewrite a periodic decimal as a fraction.

That is, the technique amounts to a constructive PROOF of a certain mathematical proposition:
namely that any periodic decimal can be written as a fraction between two natural numbers, i.e a periodic decimal is necessarily a rational number.
 
100x - x=99x=91

x=91/99

Thats why its used, to get urself a fraction.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top