Project Euler Problem 26: Reciprocal cycles
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1

Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

Find the value of d<1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

 '''
 Author: Walker Kroubalkian
 Order of 10 Approach to Project Euler Problem #26
 '''

 import time

 def projectEulerProblemTwentySix(n):
     maxIndex = 1
     maxCycle = 0
     for i in range(2,n):
         temp = i
         while(temp%2==0):
             temp/=2
         while(temp%5==0):
             temp/=5
         if(temp!=1):
             e = 1
             p = 10
             while(p%temp!=1):
                 p*=10
                 p%=temp
                 e+=1
             if(e>maxCycle):
                 maxIndex = i
                 maxCycle = e
     return maxIndex

 start = time.time()
 print projectEulerProblemTwentySix(1000)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 983
 --- 0.0106329917908 seconds ---

 for input of n=1000
 '''

Project Euler Problem 24: Lexicographic permutations
A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1, and 2 are:

012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9?

 '''
 Author: Walker Kroubalkian
 Brute Force Approach to Project Euler Problem #24
 '''

 import time

 def factorial(n):
     total = 1
     for c in range(2,n+1):
         total*=c
     return total

 def projectEulerProblemTwentyFour(n, x):
     myFactorials = []
     for i in range(x):
         myFactorials.append(factorial(x-1-i))

     myString = ""
     n-=1
     counter = 0

     possibleDigits = []
     for i in range(x):
         possibleDigits.append(str(i))

     while(counter<x and len(possibleDigits)>1):
         total = 0
         thisDigit = -1
         while(total<=n):
             total+=myFactorials[counter]
             thisDigit+=1
         myString += possibleDigits[thisDigit]
         n-=(total-myFactorials[counter])
         possibleDigits.pop(thisDigit)
         counter+=1

     myString+=possibleDigits[0]

     return myString

 start = time.time()
 print projectEulerProblemTwentyFour(1000000, 10)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 2783915460
 --- 2.90870666504e-05 seconds ---

 for input of n = 1000000, x = 10
 '''

Project Euler Problem 23: Non-abundant sums
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

 '''
 Author: Walker Kroubalkian
 Sieve Approach to Project Euler Problem #23
 '''

 import time

 def projectEulerProblemTwentyThree(n):
     sums = [0]*(n+1)
     for i in range(2,n/2+1):
         for c in range(2*i, n+1, i):
             sums[c]+=i

     abundant = []

     for i in range(n+1):
         if(sums[i]>i):
             abundant.append(i)

     possible = [True]*(n+1)
     possible[0] = False

     for a in abundant:
         for b in abundant:
             if(b>a or (a+b) > n):
                 break
             possible[a+b] = False

     total = 0

     for i in range(n+1):
         if(possible[i]):
             total+=i

     return total

 start = time.time()
 print projectEulerProblemTwentyThree(28123)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 4179871
 --- 0.863159894943 seconds ---

 for input of n=28123
 '''

Project Euler Problem 22: Names scores
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 x 53 = 49714.

What is the total of all the name scores in the file?

 '''
 Author: Walker Kroubalkian
 Brute Force Approach to Project Euler Problem #22
 '''

 import time

 f = open("PE22Names.txt","r")

 if(f.mode == "r"):
     contents = f.readlines()
     realContents = []
     for x in contents:
         realContents.append(x.split(","))
 else:
     raise ValueError("Cannot read from file")

 finalList = realContents[0]

 def getValueString(myString):
     total = 0
     for x in myString:
         if("A"<=x<="Z"):
             total+=(ord(x)-ord("A")+1)
     return total

 def projectEulerProblemTwentyTwo(nameList):
     nameList = sorted(nameList)
     total = 0
     numberNames = len(nameList)
     for i in range(numberNames):
         total+=(i+1)*getValueString(nameList[i])
     return total

 start = time.time()
 print projectEulerProblemTwentyTwo(finalList)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 871198282
 --- 0.00785303115845 seconds ---

 for input of given list of names
 '''

Project Euler Problem 21: Amicable numbers
Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71, and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

 '''
 Author: Walker Kroubalkian
 Semi-Sieve Approach to Project Euler Problem #21
 '''

 import time
 from math import sqrt

 def addProperDivisors(n):
     temp = n
     p = []
     e = []
     c = 2

     while(c<=sqrt(n)):
         if(n%c==0):
             t = 0
             while(n%c==0):
                 n/=c
                 t+=1
             p.append(c)
             e.append(t)
         c+=1

     if(n>1):
         p.append(n)
         e.append(1)

     total = 1
     powers = len(p)

     for x in range(powers):
         powerSum = 0
         for i in range(e[x]+1):
             powerSum+=(p[x]**i)
         total*=powerSum

     return total - temp

 def projectEulerProblemTwentyOne(n):
     sums = [0]*n
     i = 1

     while(i<=n/2):
         for c in range(2*i, n, i):
            sums[c]+=i
         i+=1

     total = 0
     potential = []

     for x in range(n):
         value = sums[x]
         if(value>=n):
             potential.append(x)
         elif(value>0 and value!=x and sums[value] == x):
             total+=x

     for x in potential:
         if(addProperDivisors(addProperDivisors(x)) == x):
             total+=x

     return total

 start = time.time()
 print projectEulerProblemTwentyOne(10000)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 31626
 --- 0.0148520469666 seconds ---

 for input of n=10000
 '''

Project Euler Problem 20: Factorial digit sum
n! means n x (n-1) x . . . x 3 x 2 x 1

For example, 10! = 10 x 9 x . . . x 3 x 2 x 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

 '''
 Author: Walker Kroubalkian
 Brute Force Approach to Project Euler Problem #20
 '''

 import time

 def projectEulerProblemTwenty(n):
     total = 1
     for i in range(2,n+1):
         if(total%10==0):
             total/=10
         total*=i
     a = str(total)
     final = 0
     for l in a:
         final+=int(l)
     return final

 start = time.time()
 print projectEulerProblemTwenty(100)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 648
 --- 0.000115871429443 seconds ---

 for input of n=100
 '''

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.

Thirty days has September,
April, June and November,
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine,
And on leap years, twenty-nine.

A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

 '''
 Author: Walker Kroubalkian
 Library Approach to Project Euler Problem #19
 '''

 import datetime
 import time

 def projectEulerProblemNineteen(m1, d1, y1, m2, d2, y2):
     start = datetime.datetime(y1, m1, d1)
     end = datetime.datetime(y2, m2, d2)
     day = datetime.timedelta(days = 1)
     week = datetime.timedelta(days = 7)

     while(start.weekday()!=6 and start<=end):
         start+=day

     total = 0
     while(start<=end):
         if start.day==1:
             total+=1
         start+=week

     return total

 start = time.time()
 print projectEulerProblemNineteen(1, 1, 1901, 12, 31, 2000)
 print ("--- %s seconds ---" % (time.time()-start))

 '''
 Prints

 171
 --- 0.000821828842163 seconds ---

 for input of start date = January 1, 1901, end date = December 31, 2000 
 '''