jakedobkin

# set up prime sieve to n
n=10000
myPrimes = [True]*(n+1)
last = 0
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i
		while j<=n:
			myPrimes[j]=False

jakedobkin

# http://projecteuler.net/problem=61
# this should run in 500ms

import itertools

# these are the formulas for the six polygonal numbers
def fn(n):
  return [n*(n+1)/2,n*n,n*(3*n-1)/2,n*(2*n-1),n*(5*n-3)/2,n*(3*n-2)]

# this is what we'll use for the cyclic pattern matching

jakedobkin

# http://projecteuler.net/problem=62

def cube(n):
	return n*n*n

def hash(n):
	string = str(n)
	numbers = []
	for i in range(0,len(string)):
		numbers.append(string[i:i+1])

jakedobkin

# pretty straightforward- but i'll leave the math here to you

count = 0
for a in range (1,10):
	for b in range (1,30):
		if len(str(a**b))==b:
			count+=1
print count

jakedobkin

import math

squares=[]
for i in range (1,10000):
	squares.append(i*i)

def CF_of_sqrt(n):
# modified this from http://eli.thegreenplace.net/2009/06/19/project-euler-problem-66-and-continued-fractions/
# who got it from http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfINTRO.html#sqrtalgsalg
    if n in squares: 

jakedobkin

# pattern in convergents of e
# 1 + 2 * 1 = 3
# 2 + 3 * 2 = 8
# 3 + 8 * 1 = 11
# new term = two terms ago + one term ago * i

def make_e(n):
	CF = [2]
	c = 1
	while len(CF)<n:

jakedobkin

# http://projecteuler.net/problem=66
# this is kind of a bear- it combines all the continued fraction stuff
# but it runs in 147ms

import math

squares=[]
for i in range (1,101):
	squares.append(i*i)

jakedobkin

# http://projecteuler.net/problem=67
# i wrote problem 18 in javascript- http://projecteuler.net/index.php?section=problems&id=18, https://gist.github.com/1380694, so i rewrote it here in python

file = open('triangle.txt','r').readlines()

# this opens the file, cleans it of the line breaks, and then imports it into a 2D array
for i in range (0,len(file)):
	file[i]=file[i][0:-2]
array=[]
for j in range (0,len(file)):

jakedobkin

# http://projecteuler.net/problem=69

# first set up a special sieve- factors[j] is the totient for each number
factors = [1]*1000001
n=1000000
myPrimes = [True]*(n+1)
last = 0
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i

jakedobkin

factors = [1]*10000001
for i in range (0,len(factors)):
	factors[i]=i


# set up prime sieve to n
n=10000000
myPrimes = [True]*(n+1)
last = 0
for i in range (2,n):