jakedobkin

# http://projecteuler.net/problem=58
# this is the same spiral from http://projecteuler.net/problem=28

# doing individual prime tests is faster than using a seive for numbers this large
def is_prime(n):
  if n == 2 or n == 3: return True
  if n < 2 or n%2 == 0: return False
  if n < 9: return True
  if n%3 == 0: return False
  r = int(n**.5)

jakedobkin

# http://projecteuler.net/problem=58

# i found this confusing, so i worked through dreamshire's analysis
# numerator is always = numerator + denominator * 2
# denominator is = numerator + denominator 
# starting with numerator 3 and denominator 2
# which is round 1, so start at round 2

numerator = 3
denominator = 2

jakedobkin

# http://projecteuler.net/problem=57

# python is well suited to doing this through simple brute force
max = 0
for a in range (1,100):
	for b in range (1,100):
		a2b = str(a**b)
		sum = 0
		for x in range (0,len(a2b)):
			digit = int(a2b[x:x+1])

jakedobkin

# http://projecteuler.net/problem=56

# first, a few functions we'll use- i particularly like the reverse

def reverse(string):
	string = string[::-1]
	return string

def is_pali(string):
 	if string==reverse(string):

jakedobkin

# arrays for grading hands

numbers = ["2","3","4","5","6","7","8","9","T","J","Q","K","A"] 
plays = ["0P","1P","2P","3K","4$","5F","6F","7F","8F","9F"]

# some functions we'll use- the basic approach for the grade function 
# is to give each hand a string grade- when sorted, the better hand should come out on top

def l2n(letter):
	index = numbers.index(letter)+2

jakedobkin

# http://projecteuler.net/problem=53
# python has a lot of useful math functions!

import math

# it also has itertools, which can give us the exact combinations of a set
# but here we're just calculating numbers, so i'll use the math factorial function

count = 0
for n in range (2,101):

jakedobkin

# http://projecteuler.net/problem=52

# we'll use the python built-in permutation function
import itertools

# 166K = 1MM/6, so that's the largest number under 1MM that could have identical # of digits 
for k in range (100000,166666):
	my_set = set()
	perms = itertools.permutations(str(k), len(str(k)))
	for i in perms:

jakedobkin

# set up prime sieve to n- we'll need up to 999999 here, since we're investigating 6 figure combos
total = 0
n=1000000
myPrimes = [True]*(n+1)
mySum = [0]*(n+1)
last = 0
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i
		while j<=n:

jakedobkin

# set up prime sieve to n- track running sum in separate sum array
total = 0
n=1000000
myPrimes = [True]*(n+1)
mySum = [0]*(n+1)
last = 0
for i in range (2,n):
	if myPrimes[i] == True:
		j = 2*i
		while j<=n:

jakedobkin

# http://projecteuler.net/problem=49

# we'll use the python built-in permutation function
import itertools

# prime sieve
n=100000
myPrimes = [True]*(n+1)
for i in range (2,n):
	if myPrimes[i] == True: