 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Scarce primes (Posted on 2011-01-10) A repunit is a number consisting solely of ones (such as 11 or 11111).
Let us call p(n) a 10-base integer represented by a string of n ones, e.g. p(1)=1, p(5)=11111 etc.
Most of the repunit numbers are composite.
2, 19,23,317 are the first four indices of prime repunits.

Prove: For a prime repunit p(n) to be prime, n has to be prime.

 No Solution Yet Submitted by Ady TZIDON Rating: 3.6667 (3 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Another proof | Comment 4 of 10 | As in an earlier comment, for n>1 the contrapositive "If n is composite, then p(n) is composite" is easier to show. (For n=1, note p(1) is not prime.)

Suppose n composite with a factorization xy, x>1, y>1. Then let q=p(y) with x-1 zeroes placed between each pair of adjacent ones, and note p(n)=p(x)*q. x>1 implies p(x)>1, and y>1 implies q>p(y)>1. Thus p(n) is composite.

An example is n=6. If x=2, y=3 then p(x)=11, q=10101.
 Posted by Gamer on 2011-01-10 18:45:40 Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information