Find the prime factorization of the integers 1234, 10140, and 36000?

AI Thread Summary
The prime factorization of 1234 is correctly stated as 2·617. For 10140, the canonical form is 2²·3·5·13², which is a more concise representation. The factorization of 36000 is accurately expressed as 2⁵·3²·5³. The discussion confirms that the provided factorizations are correct and suggests using WolframAlpha for verification. Overall, the thread emphasizes the importance of presenting prime factorizations in canonical form.
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Homework Statement
Find the prime factorization of the integers ## 1234, 10140 ##, and ## 36000 ##.
Relevant Equations
None.
## 1234=2\cdot 617 ##
## 10140=2\cdot 2\cdot 3\cdot 5\cdot 13\cdot 13 ##
## 36000=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5\cdot ##

Are the answers above correct? Or do I need to put them in canonical form as below?

## 1234=2\cdot 617 ##
## 10140=2^{2}\cdot 3\cdot 5\cdot 13^{2} ##
## 36000=2^{5}\cdot 3^{2}\cdot 5^{3} ##
 
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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
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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