Taylor Series approach (alternative solution)
I was able to solve this without continued fractions for 100pts. I’ll share the approach I commented on another question, since it seems relevant. Alternatively, we can also use Taylor Series of e:
e=\sum_{k=0}^\infty\frac{1}{k!}
For a good approximation, around M=3000 iterations is enough, and you can simplify:
e\approx\sum_{k=0}^M\frac{1}{k!}=\frac{\sum_{k=0}^M(M!/k!)}{M!}
I stored the numerator and denominator as big integers, and got accepted with python with similar summing floor trick found at part 1 above (*). Unfortunately, I am not able prove why M=3000 is enough, I would appreciate any comments. I tried to do the usual taylor approximation (which resulted to M=1465, but gave WA), so I would like some help in finding a better approximation method for taylor series with summing floor. Here’s a link to my solution 
