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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)

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Solution proof | Comment 5 of 10 |
Suppose n is composite, and equals a*b (a and b both > 1)

Then p(n) can be subdivided into a groups of b digits each. Each group is p(b) times an appropriate power of 10, so p(n) = p(b) * [1 + 10^b + 10^2b + ... + 10^(a-1)b] (there are a terms in the second factor)

So, p(n) for composite n is also composite unless either p(b) or the second factor is 1. Well, p(b) can't be 1 because b > 1 (see above) and the second factor is also only 1 if a = 1, which also contradicts the initial assumption.)

So all p(n) when n is composite are composite.

That doesn't mean that p(n) is necessarily prime when n is prime, but it does mean there are no cases where n is composite and p(n) is prime, and hence that for p(n) to be prime, n must also be prime.

  Posted by Paul on 2011-01-10 22:23:15
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