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Old 2004-04-26, 03:05   #6
Maybeso
 
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Aug 2002
Portland, OR USA

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Using brute force on the end 5dpf's:
~ 50000 odds less than 100000
~ 40000 with 5's removed, there should be
~ 20000 pairs, but we lose those matched with the missing 5's, so actual pairs
= 11145.

So the plan could be to sieve a block of k*10^5 + 45903 and k*10^5 + 46303. Removing a number from the low end of one group should also remove one from the high end of the other.

If handled properly, that will reduce the work by a factor of 7.
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