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2018-01-02, 06:44   #3
CRGreathouse

Aug 2006

22×1,493 Posts

Quote:
 Originally Posted by George M 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?
By induction. If there are 2^k divisors for the product P, and you multiply P by some prime not in P (call it q), the divisors of the new number Pq are the divisors of P, together with q times the divisors of P. Since there is no overlap (why?), there are 2^k + 2^k = 2^(k+1) divisors. Now just test the base case and you can add QED.

Quote:
 Originally Posted by George M Perhaps it could be related to Mersenne Primes?
No.