I don't think I understand your algorithm here. When trying to calculate, say 100!, what should the second argument be? I thought about this method, but I opted to try out a different method. You are correct in saying that much time is spent multiplying numbers. So what we can do is get rid of that multiplication and turn it into addition. This can be easily done by first taking logs of all integers in the factorial calculation, summing those values and finally raising e to that power. For example, 5!=e^(ln(2)*ln(2)*ln(4)*ln(5)). There is only one problem with that. Since Math::BigFloat does not contain a log or exp function, the limited precision of the built in functions creates a rounded answer (try doing 200!)
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