I’m working to bone up on my python skills so I decided to spend my Sunday doing problems 1-10 from Project Euler. I’ve done them before with C or Java but this was my first time with Python. Here are the problems and my commented code for each one in case it interests anybody.

### Problem 1 – Multiples of 3 and 5

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. **ANSWER = 233168**.

# Project Euler - Question 1 - Multiples of 3 & 5 # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=1 # If we list all the natural numbers below 10 that are # multiples of 3 or 5, we get 3, 5, 6 and 9. # The sum of these multiples is 23. Find the sum of # all the multiples of 3 or 5 below 1000. ANSWER = 2318. sum = 0 #variable to hold sum # Iterate i from 0 to 99 # If i is divisible by 3 or 5, add to sum for i in range(100): if (i%3 == 0 or i%5==0): sum += i # Print answer print('The sum is: ' + str(sum))

### Problem 2 – Even Fibonacci Numbers

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. **ANSWER = 4613732**

# Project Euler - Question 2 - Even Fibonacci Numbers # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=2 # Each new term in the Fibonacci sequence is generated by # adding the previous two terms. By starting with 1 and 2, # the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # By considering the terms in the Fibonacci sequence whose values do # not exceed four million, find the sum of the even-valued terms. # Answer = 4613732 sum = 0 # Variable to hold sum num1 = 0; # First number num2 = 1; # Second number # While the second number is less than 4000000 # This ensures the first number is less after moving while num2 < 4e6: # Method 1 of incrementing numbers temp = num1 num1 = num2 num2 = num1 + temp # Method 2: # num1, num2 = num2, num1+num2 # If the number is eve, add to sum if(num1%2 == 0): sum += num1 # Print results print('The sum is: ' + str(sum))

### Problem 3 – Largest Prime Factor

The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143? **ANSWER = 6857**

# Project Euler - Question 3 - Largest Prime Factor # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=3 # The prime factors of 13195 are 5, 7, 13 and 29. # What is the largest prime factor of the number 600851475143? # Answer = 6857 # Import math library to get sqrt import math # isPrime function - returns True or False def isPrime(num): # Iterates from 2 to sqrt(num)+1 as discussed in #7 # Make sure to convert sqrt to int for range # Using xrange will save considerable time for large numbers for i in xrange(2,int(math.sqrt(num))+1): if (num % i == 0): return False return True # List with factors, starts with only one factors = [600851475143] # Infinite loop, will break out when necessary while True: # If the largest factor (0 spot) is prime, break if(isPrime(factors[0])): break # Try to divide the largest factor by numbers starting with 2 # If a number evenly divide, reduct the largest factor and # append the divisor to the end of the list, break out of for # loop and start again at 2 with the new largest factor for i in xrange(2,factors[0]): if (factors[0] % i == 0): factors[0] = factors[0] / i factors.append(i) break # Sort factors and print them factors.sort() print(factors)

### Problem 4 – Largest Palindrome Product

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers. **ANSWER = 906609**

# Project Euler - Question 4 - Largest Palindromic Product # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=4 # A palindromic number reads the same both ways. # The largest palindrome made from the product # of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of two 3-digit numbers. # ANSWER: 913 * 993 = 906609. Took 1.0s # isPal function returns True or False # Converts number to string and iterates through half of it # If characters don't match return false def isPal(num): numString = str(num) for i in range(0,len(numString)/2+1): if (numString[i] != numString[-i-1]): return False return True # Keep track of max product maxProduct = 0 # Keep track of values used for max prodcut max1, max2 = 0,0 # Iterate from 999 to 99 going down for both values # Generate a product and determine if it's a palindrome # If it is a palindrom, determine if it is max for i in range(999, 99, -1): for j in range(999, 99, -1): product = i * j if isPal(product): if(product > maxProduct): maxProduct = product max1, max2 = i, j # Print snswers print('The largest palindromic product is: ' + str(maxProduct)) print(str(maxProduct) + ' = ' + str(max1) + ' * ' + str(max2))

### Problem 5 – Smallest Multiple

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? **ANSWER = 232,792,560**

# Project Euler - Question 5 - Smallest Multiple # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? # Answer = 232,792,560 # First time I wrote this I started from 1 going up and # divided by all the numbers from 1 up to 20 and counted # number of multiples. If it got to 20 it stopped and printed # Took about (290) to run. # Second time I wrote it, I started from 1 going up and # divided by all the numbers from 20 down to 1 and counted # number of multiples and if I got to 20 it stopped and printed # Took about 224 seconds to run. This speeds it up because a number # is less likely to be divisible by 20 or 19 than by 1 or 2 so by starting # at the top you rule out that number quicker and move on. # Third time I wrote it, I started from 1 going up and # divided by all the numbers from 20 down to 11 and counted # number of multiples and if I got to 10 it stopped and printed # Took about 210 seconds to run. This speeds it up because if a number # is divisible by 20, it's also divisible by 10, by 18 it's also by 9, # by 16, it's also by 8, etc. So we don't need to test for 1-10 # Forth time I wrote it, I started from 20 going up by 20 and # divided by all the numbers from 20 down to 11 and counted # number of multiples and if I got to 10 it stopped and printed # Took about 14 seconds to run. We can do this because all of these # numbers must be divisible by 20 so checking anything between the multiples # of twenty is a waste of time. # Fifth time I wrote it, I started from 2520 going up by 2520 and # divided by all the numbers from 20 down to 11 and counted # number of multiples and if I got to 10 it stopped and printed # Took about 0.2 seconds to run. We can start at 2520 since the # problem tells us that is the lowest number for 1-10 multiples # and since 1-20 includes 1-10, it can't be less. We can increase # by 2520 for the same reason. # The code below is from the forth time I wrote it with two # solutions using different techniques # ------------------------------------------------------ # First solution -------------------------------------- # ------------------------------------------------------ num = 0 # Infinite Loop because we don't know the max # We will use a break when we find the first number # Could also do a for loop up to 20! but this is easier while True: #increment num by 20 each time num += 20 # NumMultiples keeps track of the number of multiples for num numMultiples = 0 # Iterate from 20 to 11 going down by 1 for i in range(20,10, -1): # If the number is divisible by i then increase # the number of multiples num has. if (num % i == 0): numMultiples+=1 else: # If not evenly divisible, break to save time break; # If there were 10 multiples that means it's evenly # divisibly by all the numbers so we break out of while if (numMultiples == 10): break; # Print out current num to keep track (in millions) if num % 1000000 == 0: print(num/1000000) print('The lowest common multiple is: ' + str(num)) # ------------------------------------------------------ # Second solution -------------------------------------- # ------------------------------------------------------ num = 0 keepGoing = True # In this version, we go until the boolean keepGoing is changed to False while keepGoing: #increment num by 20 each time num += 20 # NumMultiples keeps track of the number of multiples for num numMultiples = 0 # Iterate from 20 to 11 going down by 1 for i in range(20,10, -1): if (num % i != 0): # If not evenly divisible, break out of for loop # because we know it isn't the LCM break if (i == 11): # If we reached here without breaking, then we've # found the number we're looking for. # Change keepGoing to False to break out of while loop keepGoing = False break # Print out current num to keep track (in millions) if num % 1000000 == 0: print(num/1000000) print('The lowest common multiple is: ' + str(num))

### Problem 6 – Sum Square Difference

The sum of the squares of the first ten natural numbers is, *1 ^{2} + 2^{2} + … + 10^{2} = 385* The square of the sum of the first ten natural numbers is,

*(1 + 2 + … + 10)*Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

^{2}= 55^{2}= 3025**ANSWER = 25164150**

# Project Euler - Question 6 - Sum Square Difference # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=6 # The sum of the squares of the first ten natural numbers is, # 1^2 + 2^2 + ... + 10^2 = 385 # The square of the sum of the first ten natural numbers is, # (1 + 2 + ... + 10)^2 = 55^2 = 3025 # Hence the difference between the sum of the squares of the first # ten natural numbers and the square of the sum is 3025 - 385 = 2640. # Find the difference between the sum of the squares of the # first one hundred natural numbers and the square of the sum. # ANSWER = 25,164,150 # Keep track of the sum of the squares sumSquare = 0 # Keep track of the the sums sums = 0 # Iterate through numbers [1:100] 1-100 inclusevely for i in range(1,101): sumSquare += i*i sums += i # Square the individuals sums to find square of sums squareSum = sums * sums # Print answers print('The sum of the squares is: ' + str(sumSquare)) print('The square of the sums is: ' + str(sums)) print('The difference is: ' + str(squareSum - sumSquare))

### Problem 7 – 10001st prime

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10,001st prime number? **ANSWER = 104,743**

# Project Euler - Question 7 - 10,001st prime # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=7 # By listing the first six prime numbers: # 2, 3, 5, 7, 11, and 13, # we can see that the 6th prime is 13. # What is the 10,001st prime number? # ANSWER = 104,743 # Took 102s in first version. # In the first version, I checked for primes from # two to num. # In the second version, I checked for primes from # two to num/2+1. # In the third version, I checked for primes from # two to sqrt(num)+1. # Version 1: 102s # Version 2: 44s # Version 3: 0.5s # Import math library to get sqrt import math # isPrime function - returns True or False def isPrime(num): # Iterates from 2 to sqrt(num)+1 as discussed above # Make sure to convert sqrt to int for range for i in range(2,int(math.sqrt(num))+1): if (num % i == 0): return False return True count = 1 # Number of primes num = 2 # Prime number (count) # While loop to continue until we reach 100001th prime while (count<10001): num+=1 if isPrime(num): #print('Found a prime: ' + str(num) + ', prime number: ' + str(count)) # If a prime is found, increase count and continue count+=1 # Print answer print('The 10001th prime is: ' + str(num))

### Problem 8 – Largest Product in a Series

The four adjacent digits in the 1000-digit number (below) that have the greatest product are 9 × 9 × 8 × 9 = 5832. Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product? **ANSWER = 23,514,624,000**

7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450

# Project Euler - Question 8 - Largest Product in a Series # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=8 # The four adjacent digits in the 1000-digit number that # have the greatest product are 9 x 9 x 8 x 9 = 5832. # 73167176531330624919225119674426574742355349194934 # 96983520312774506326239578318016984801869478851843 # 85861560789112949495459501737958331952853208805511 # 12540698747158523863050715693290963295227443043557 # 66896648950445244523161731856403098711121722383113 # 62229893423380308135336276614282806444486645238749 # 30358907296290491560440772390713810515859307960866 # 70172427121883998797908792274921901699720888093776 # 65727333001053367881220235421809751254540594752243 # 52584907711670556013604839586446706324415722155397 # 53697817977846174064955149290862569321978468622482 # 83972241375657056057490261407972968652414535100474 # 82166370484403199890008895243450658541227588666881 # 16427171479924442928230863465674813919123162824586 # 17866458359124566529476545682848912883142607690042 # 24219022671055626321111109370544217506941658960408 # 07198403850962455444362981230987879927244284909188 # 84580156166097919133875499200524063689912560717606 # 05886116467109405077541002256983155200055935729725 # 71636269561882670428252483600823257530420752963450 # Find the thirteen adjacent digits in the 1000-digit # number that have the greatest product. # What is the value of this product? # ANSWER = 23,514,624,000. Took 0.1s # findProduct function takes a string and returns # the product of every digit def findProduct(numString): product = 1 for i in range(len(numString)): product = product * int(numString[i]) return product # The number we need in int and string form bigNum = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450 bigString = str(bigNum) # Variables to track things maxProduct = 0 product = 0 numChars = 13 # Iterate over the 1000 digits except for the end digits # For each iteration find the string of numChar length and # pass it to the function to find product for i in range(len(bigString)-numChars): product = findProduct(bigString[i:i+numChars]) # If a product is larger than max, store it if (product > maxProduct): maxProduct = product # Print answers print('Max Product of ' + str(numChars) + ' digits is: ' + str(maxProduct))

### Problem 9 – Special Pythagorean Triplet

A Pythagorean triplet is a set of three natural numbers, `a` < `b` < `c`, for which,

`a`

^{2}+

`b`

^{2}=

`c`

^{2}

For example, 3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}.

There exists exactly one Pythagorean triplet for which `a` + `b` + `c` = 1000. Find the product `abc`. **ANSWER = 31,875,000**

# Project Euler - Question 9 - Special Pythagorean Triplet # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=9 # A Pythagorean triplet is a set of three natural numbers, # a < b < c, for which, a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. # Answer = 31,875,000 Took 21 seconds. (10s with exit()) # Iterate a from 1 to 1000 # Then iterate b from a to 1000 # Then iterate c from b to 1000 # This is because a < b < c. Otherwise it would take # much much longer and return two answers for a in range(1,1000): for b in range(a,1000): for c in range (b, 1000): # Once we have an iteration of a, b, and c # Determine if it fits the criteria of a+b+c==1000 and # a^2 + b^2 == c^2 # Test a+b+c first because it is a faster test. Takes ~half the time if (a+b+c == 1000): if (a*a + b*b == c*c): # Print answers print('A: ' + str(a) + ' B: ' + str(b) + ' C: ' + str(c)) print('Product is: ' + str(a*b*c)) # If we found it, exit to save time exit()

### Problem 10 – Summation of Primes

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. **ANSWER = 142,913,828,922**

# Project Euler - Question 10 - Summation of Primes # Written by Matthew Walker, 20 August 2017 # https://projecteuler.net/problem=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. # ANSWER = 142,913,828,922 Took 14 seconds. # Using xrange in isPrime function saves considerable time. # 29s using just range, 20s using range but omitting odds # 14s using xrange, 13.5s using xrange and omitting odds # Import math library to get sqrt import math # isPrime function - returns True or False def isPrime(num): # Iterates from 2 to sqrt(num)+1 as discussed in #7 # Make sure to convert sqrt to int for range # Using xrange will save considerable time for large numbers for i in xrange(2,int(math.sqrt(num))+1): if (num % i == 0): return False return True sum = 0 # Iterate from 2 to two million # You can increase speed slightly by starting at 3 and # iterating by 2 to skip evens but only saves 0.5s for i in range(2,2000000): # If number is prime, add to sum if isPrime(i): sum += i # Print out results print('The sum of all primes below 2 million is: ' + str(sum))