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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?

Note by Gabriel Kong
4 years, 3 months ago

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You can review Divisors of an Integer. Staff · 4 years, 3 months 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 $$d(n)$$. · 4 years, 3 months ago

Thanks a lot! :D · 4 years, 3 months ago

What shourya pointed out was the number FACTORS of a number, not the number of PRIME factors. · 4 years, 3 months ago

http://en.wikipedia.org/wiki/Divisor_function and http://mathworld.wolfram.com/Divisor.html · 4 years, 3 months ago