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Old 2003-05-29, 07:47   #7
1260
 
Feb 2003

25 Posts
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Okay. If it is really THAT LARGE, isn't there a mathematical conversion to do something similar?

I mean, if S(n-1) = A mod M(n) = B mod M(n+k), then computation can start from S(n-1) to S(n-1+k) for M(n+k) which would be shorter. Is it then possible to solve for B from A, M(n) and M(n+k)?

If not, maybe more values might help:
if S = A mod M(n+a) = B mod M(n+b) = C mod M(n+c) = D mod M(n+d), where a<b<c<d, can D be computed from A, B, and C?

I'm not really familiar with all the theorems in modular arithmetic involving a change of modulo. Is there one related to this problem?

By the way, is the remainder of S(n-1) mod M(n) saved in the results.txt file? Which is it, the Res64 or the WY1?
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