ebartrum

#Define the function f by recursion:
def f(n):
    if n==1:
        return 1
    if n==3:
        return 3
    if n%2==0:
        return f(n/2)
    if (n-1)%4==0:
        m=(n-1)/4

ebartrum

#Define a function to return true if we have a digit cancelling fraction
 
"""The function will take 2 inputs: [a,b],[c,d].
It will return true if ab/cd is a digit cancelling fraction."""
 
#Define a function to return a common element of x,y
 
def commonElt(x,y):
    for z in x:
        if z in y:

ebartrum

#Define a function pyth(a,b,c) which returns true if a^2+b^2=c^2


def pyth(a,b,c):
    if a**2+b**2==c**2:
        return True
    else:
        return False
#first check (1,x,y)

ebartrum

x="73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\

ebartrum

#Define a function which checks if any members of x divide y (If so return True, otherwise False) but only checks up to the squareroot of y (This greatly improves efficiency).
#In other words if the function finds a member of x that is >= the sqrt of y it stops looking and returns the current state (ans)
def divis(x,y):
    ans=False
    for z in x:
        if y%z==0:
            ans=True
            break
        if z>=y**0.5+1:
            break

ebartrum

import time

#Define a function which checks if any members of x divide y (If so return True, otherwise False) but only checks up to the squareroot of y (This greatly improves efficiency).
#In other words if the function finds a member of x that is >= the sqrt of y it stops looking and returns the current state (ans)
def divis(x,y):
    ans=False
    for z in x:
        if y%z==0:
            ans=True
            break

ebartrum

ans=1
n=1001
y=range(3,n+1,2)
#The solution comes from the observation that the sum of 4 nxn corner squares is the following poly (for odd n>1)
def poly(x):
    ans=4*x**2-6*x+6
    return ans
for o in y:
    ans+=poly(o)
print ans  

ebartrum

"""I wish to create a function lcm(x) which gives the lowest common multiple of an array of numbers x"""

# I will need the factor function

def factor(x):
    array=[]
    i=1
    while (i<=x):
        if x%i==0:
            array.append(i)

ebartrum

"""A palindromic number reads the same both ways. Find the
largest palindrome made from the product of two 3-digit numbers."""

#Define a 'surround' function that will be used to construct palindromes
def surround(x,y):
    x=str(x)
    y=str(y)
    return str((y+x+y))

#start with an initial array of all palidromes of length 1 or 2

ebartrum

"""The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?"""

#I want to define prime(x) which gives x's prime factors

#Define factor(x)

def factor(x):
    array=[]