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Topic: Same Number of Divisors (Read 2638 times) |
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Barukh
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Same Number of Divisors
« on: Oct 5th, 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
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Re: Same Number of Divisors
« Reply #1 on: Oct 5th, 2007, 10:31am » |
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on Oct 5th, 2007, 7:58am, Barukh wrote:
This one is 4, unless | hidden: | 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. |
There are no such solutions below 1016, 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 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. |
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| « Last Edit: Oct 5th, 2007, 2:28pm by Eigenray » |
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