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        riddles >> medium >> Same Number of Divisors
        
(Message started by: Barukh on Oct 5^th, 2007, 7:58am)

Title: Same Number of Divisors
Post by Barukh on Oct 5^th, 2007, 7:58am

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?

Title: Re: Same Number of Divisors
Post by Eigenray on Oct 5^th, 2007, 10:31am

on 10/05/07 at 07:58:11, Barukh wrote:

b) N = 6

This one is [hide]4[/hide], unless [hideb]there is some n with six divisors such that

n+1 = 2p²,
n+2 = 3r² or 3²r,
n+3 = 4q,
n+4 = 5s² or 5²s

for primes p,q,r,s.[/hideb]
There are no such solutions below 10¹⁶, at least.

Edit: Hah! Letting my program run a few more seconds:
[hide]{10093613546512321, 10093613546512322, 10093613546512323, 10093613546512324, 10093613546512325}[/hide]
Here's a few more:
[hide]{266667848769941521, 266667848769941522, 266667848769941523, 266667848769941524, 266667848769941525}
{1579571757660876721, 1579571757660876722, 1579571757660876723, 1579571757660876724, 1579571757660876725}[/hide]

Proof that these have maximal length:[hideb]Each number must be of the form p⁵ or p²q. If there are 6 consecutive numbers, one of them must be divisible by 6, and could only be either 2²*3 or 2*3², but both of these are surrounded by primes.[/hideb]