> From: Brian J. Beesley [mailto:[EMAIL PROTECTED]]
> What I mean is that 10^100 = 3 + (10^100 - 3) is one possible 
> breakdown. But I don't know offhand whether (10^100 - 3) is prime, 
> and proving that it is might take some considerable time.  If that

However, proving it is composite is very easy.  It is divisible by 13.

> fails, I can skip 5 + (10^100 - 5) and 7 + (10^100 - 7) since in each 
> case the larger number is clearly composite. But, in general, I'd 
> have to check whether (10^100 - p) is prime for every odd prime p 
> until I find a prime (proving that Goldbach's Conjecture holds for 
> the specific odd number 10^100) or until p > 0.5 * 10^100 (a 
> counterexample!)

1.96 seconds on a PII-300 and Francois Morain's ECPP program shows that
10^100-797 is prime.  Running a composite test on all the other larger
candidates took 1.80 seconds in total.


Paul
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