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Old 2018-01-02, 05:55   #1
George M
Dec 2017

2·52 Posts
Lightbulb 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.
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