# Euler Solution 65

### From ProgSoc Wiki

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

# Solutions for Problem 65

## Python by Althalus

Runtime: 4ms

limit=100 #e = [2; 1,2,1, 1,4,1, 1,6,1, ..., 1,2k,1...] #Let's start off by gettingthe first 99 parts of our repeated fraction. es = [2,] count=1 #our first convergent won't be in our list es. while len(es) < limit: es+=[1,2*count,1] count+=1 es.reverse() fraction = [1,es[0]] #evaluate.... for e in es[1:]: fraction[0],fraction[1]=fraction[1],fraction[0] fraction[0]+=e*fraction[1] print sum([int(x) for x in str(fraction[0])])