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 \(d(n)\).

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TopNewestYou can review Divisors of an Integer.

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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 \(d(n)\).

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Thanks a lot! :D

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

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

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wow! thanks.

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