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 Minimum Value Muse (Posted on 2014-06-20)
Determine the minimum value of a positive integer N such that each of N, N+1, N+2, and N+3 has the same number of divisors (including 1 and itself.)

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible solution with thought process | Comment 1 of 3
An idea that seemed to work:

Too small a number of divisors doesn't seem likely.

What if there are 6 divisors?  The numbers could be of the form p*q^2.
For one of them p=2 and the other q=2.
These numbers would be 2 apart
So we'd have 2*q^2 = p*2^2 ±2
I then made a short list of potential candidates where both p and q are prime:  (p,q)
(5,3)
(13,5)
(61,11)
(181,19)
I stopped here thinking I'd increase my search if needed.
These pairs yield two numbers with 6 factors that are 2 apart:
18 and 20
50 and 52
242 and 244
722 and 724
Next to check if the number between has 6 factors:
19 - no (just 2)
51 - no (just 4)
243 - yes this is 3^5 (I hadn't thought of that)
723 - no (just 4)
so we have 3 numbers 242,243,244 and need a 4th at either end.
241 is prime
245 = 5*7^2 does it!

n=242

[In hindsight this could have been easier and there were a lot of pitfalls but I looked this up https://oeis.org/A006558 so this answer is right.]

 Posted by Jer on 2014-06-20 11:48:18

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