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Let $n$ be an integer with prime factorization $n = p_1^{n_1} p_2^{n_2} ... p_k^{n_k}$ , where the $p_i$ are distinct primes. How many positive divisors does the integer $n$ have?

Here is what I have:

$p_1^{n_1}$ has the following divisors: $1, p_1, {p_1}^2, {p_1}^3, ... p_1^{n_1}$

So that is $(n_1 + 1)$ divisors

And for $p_2^{n_2}$ it would be $(n_2 + 1)$, and so on.

So n has $(n_1 + 1)(n_2 + 1)(n_3 +1)... (n_j + 1)$ divisors.

Is this correct?

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    $\begingroup$ Yes, that's correct. $\endgroup$
    – Théophile
    Commented Oct 19, 2020 at 19:39
  • $\begingroup$ Well done. Only : the last factor in the product has to be $n_k+1$ , if we follow the prime factorization above. $\endgroup$
    – Peter
    Commented Oct 25, 2020 at 15:06

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