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 2018-01-02, 05:55 #1 George M   Dec 2017 2·52 Posts So how must we be able to prove the following? I made a conjecture, but don’t know how to prove it. Perhaps it could be related to Mersenne Primes? Consider $A =\{all \ divisors \ of \ x \in \mathbb{N}\}$. If $x = \prod_{m=1}^{k}p_m$ then $n(A) = 2^k$. This is to say, if a natural number (positive integer) $x$ is equal to the product of the 1st prime, the 2nd prime, the 3rd prime, and so on, until the k-th prime, then the amount of all the divisors of $x$ (including 1 and $x$) will be equal to $2^k$. How must we go about proving this? If we do, perhaps we could build an algorithm to find a prime value for k such that $2^k - 1$ is prime.