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Topic: Same Number of Divisors (Read 2583 times) 

Barukh
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Same Number of Divisors
« on: Oct 5^{th}, 2007, 7:58am » 
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1. What is the length of the maximal sequence of consecutive positive integers that have the same number N of divisiors, where: a) N = 4 b) N = 6 c) N = 8 2. Does there exist a sequence of 12 consecutive integers that have the same number of divisors?


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Eigenray
wu::riddles Moderator Uberpuzzler
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Re: Same Number of Divisors
« Reply #1 on: Oct 5^{th}, 2007, 10:31am » 
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on Oct 5^{th}, 2007, 7:58am, Barukh wrote: This one is 4, unless hidden:  there is some n with six divisors such that n+1 = 2p^{2}, n+2 = 3r^{2} or 3^{2}r, n+3 = 4q, n+4 = 5s^{2} or 5^{2}s for primes p,q,r,s.  There are no such solutions below 10^{16}, at least. Edit: Hah! Letting my program run a few more seconds: {10093613546512321, 10093613546512322, 10093613546512323, 10093613546512324, 10093613546512325} Here's a few more: {266667848769941521, 266667848769941522, 266667848769941523, 266667848769941524, 266667848769941525} {1579571757660876721, 1579571757660876722, 1579571757660876723, 1579571757660876724, 1579571757660876725} Proof that these have maximal length:hidden:  Each number must be of the form p^{5} or p^{2}q. If there are 6 consecutive numbers, one of them must be divisible by 6, and could only be either 2^{2}*3 or 2*3^{2}, but both of these are surrounded by primes. 

« Last Edit: Oct 5^{th}, 2007, 2:28pm by Eigenray » 
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