Are There Other Strange Division Patterns Besides Cancelling 6's?

AI Thread Summary
The discussion explores various division patterns where fractions simplify to the same value despite apparent complexity, specifically focusing on examples like 16/64, 166/664, and 1666/6664, all equating to 0.25. It highlights that these fractions maintain their value when both the numerator and denominator are multiplied by the same factor. The conversation also mentions similar patterns with other numbers, such as 19/95 and 26/65, suggesting a broader phenomenon in division. Participants consider the mathematical reasoning behind these patterns and the concept of canceling digits. Overall, the thread emphasizes the intriguing nature of these division relationships.
madah12
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Homework Statement




how come that 16/64=.25
166/664=.25
1666/6664=.25
and any 1then n number of sixes / the same number of sixes then 4 = .25
same thing with 19 / 95

is there other strange division patterns?
 
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(1 followed by n sixes) * 4 = (string of n sixes followed by 4)
 
\frac{166}{664} = \frac{16 * 13.75}{64 * 13.75} = \frac{16}{64} = \frac{1}{4}
\frac{1666}{6664} = \frac{16 * 104.375}{64 * 104.375} = \frac{16}{64} = \frac{1}{4}

You could probably show the same sort of thing is happening with 19/95, 199/995, and so on.

In all these examples the numerator and denominator of 16/64 are being multiplied by the same number, yielding a fraction that is equal to 16/64 = 1/4. Same with 19/95 and the others.
 
Several things like this. Consider 26/65 (or maybe even 49/98). Consider fractions like (10a+b)/(10b+c) = a/c or similar
 
Here's one I always liked-just cancel the 6's.
\frac{1\rlap{/}6}{\rlap{/}64} \;=\;\frac{1}{4}
:smile:
 
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
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