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For any natural number $n>1$, with factorization $$ n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m} $$ we can find the number of divisors by using the formula, $$ (\alpha_1+1)(\alpha_2+1)....(\alpha_m+1) $$ How do I find this formula. Is there a proof for it?

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  • $\begingroup$ Do you know multiplicative functions? $\endgroup$ – s1mple Apr 12 at 12:55
  • $\begingroup$ Try it for some small values of $\alpha$. ... say $n=6$. $\endgroup$ – Donald Splutterwit Apr 12 at 12:57
  • $\begingroup$ @s1mple how do i use them as a proof? $\endgroup$ – Kirito Apr 12 at 13:00
  • $\begingroup$ @Kirito please see the explanation. $\endgroup$ – s1mple Apr 12 at 13:02
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To prove this first consider the number of the form $n=p^{\alpha}$. The divisors are $1,p,p^2,\cdots, p^{\alpha}$, i.e. $d(p^{\alpha})=\alpha+1$. Now, consider $n=p^{\alpha}q^{\beta}$, where $p,q$ are co-prime. The divisors would be : $$1,p,p^2,\cdots p^{\alpha}$$

$$q,pq,p^2q,\cdots,p^{\alpha}q$$ Similarly, upto: $$q^{\beta},pq^{\beta},\cdots,p^{\alpha}q^{\beta}$$ In this case, $d(p^{\alpha}q^{\beta})=(\alpha+1)(\beta+1)$. Hence $d(n)$ is multiplicative function.

For any natural number $n>1$, with prime factorization , where $q_1,q_2\cdots q_m$ are distinct primes and $m\ge1 $ $$ n = q_1^{\alpha_1}q_2^{\alpha_2}q_3^{\alpha_3}....q_m^{\alpha_m} $$ we can find the number of divisors by using the formula, $$ (\alpha_1+1)(\alpha_2+1)....(\alpha_m+1) $$

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  • $\begingroup$ If this answer was helpful for you, please consider clicking the checkmark. Glad to help! $\endgroup$ – s1mple Apr 13 at 4:46
  • $\begingroup$ Exactly what I was looking for! Thank you. $\endgroup$ – K.defaoite Aug 26 at 12:29

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