Douglas K Chester NONLILLY wrote:
> [...]
I don't remember offhand which B1 was used, but for k to be B1 smooth,
it would need to be at least 73038. E would then be
2 * 3 * .... * 73037 - or 73037#
Looking at some of the largest primorial primes, I see that 42209# has
18241 digits - so I guess 73037# has over 30000 digits. And then we are
multiplying 3 by itself this number of times! IS this the case?
This is correct. In fact, E is the product of all primes and prime
powers <=B1, so for B1=100000, E = 2^16 * 3^10 * 5^7 * ... * 99991.
Then 3 gets exponentiated modulo Mp by 2*p*E, but not by multiplying 3
by itself that many times, but by a binary powering ladder. That
requires "only" about log_2(E) ~= B1/log(2) multiplication modulo Mp, or
something about 150000 for B1=100000.
Alex
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