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Scarce primes (Posted on 2011-01-10) Difficulty: 2 of 5
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.)
proof | Comment 1 of 10
p(n) = (10^n - 1)/9

The contrapositive of the statement is easier to prove:
if n is composite then p(n) is composite

let n = x*y, with integers x>1 and y>1

9p(n) = (10^xy - 1) [a string of xy 9s]
= ((10^x)^y - 1)
This is divisible by 10^x - 1  [a string of x 9s]

since 10^x - 1 is not 9, we have another factor of p(n)
thus p(n) is composite.

  Posted by Jer on 2011-01-10 15:01:43
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