Number Theory

Is there a theorem that describes how many prime factors a number has? And does this this theorem have a solution on how to find them?

#HelpMe! #Math

Easy Math Editor

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Math	Appears as
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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

You can review Divisors of an Integer.

Calvin Lin Staff - 8 years ago

If a number $N= p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{m}^{a_{m}}$ , for primes $p_{1},p_{2}\cdots p_{m}$ and natural numbers $a_{1},a_{2}\cdots a_{m}$ , then the number of factors that $N$ has is denoted by $d(N)$ and is given by the formulae: $d(N)= (a_{1}+1) (a_{2}+1)\cdots (a_{m}+1)$ .

And the proof is combinatoric. If $S= p_{1}^{b_{1}}p_{2}^{b_{2}}\cdots p_{m}^{b_{m}}$ is a factor of $N$ , where $b_{1},b_{2}\cdots b_{m}$ are natural, then each $b_{i} \leq a_{i}$ for all indices $i$ . So ecah $b_{i}$ can be chosen in $(a_{i}+1)$ ways, so the total number of such $S$ is $d(n)$ .

Shourya Pandey - 8 years ago

Thanks a lot! :D

Gabriel Kong - 8 years ago

http://en.wikipedia.org/wiki/Divisor_function and http://mathworld.wolfram.com/Divisor.html

Lab Bhattacharjee - 8 years ago

wow! thanks.

Gabriel Kong - 8 years ago

What shourya pointed out was the number FACTORS of a number, not the number of PRIME factors.

Harrison Lian - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$