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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.)
re(2): proof | Comment 7 of 10 |
(In reply to re: proof by Ady TZIDON)


The statements a→b and not-b→not-a are called contrapositives and are equivalent.  (If one is true the other is true.)

a = p(n) is prime
b = n is prime
not-a = p(n) is composite
not-b = n is composite

I proved the contrapositive and this implies the truth of the original statement, so a→b is proved.

Pointing out that R(3) is composite is a counterexample to b→a, which is the converse and may or may not be true for any given statement. 

  Posted by Jer on 2011-01-11 10:18:20

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