Euler Problems

Project description: Project Euler is a collection of interesting mathematical problems that are useful for learning and applying math theorems to programming. Using the techniques from Clean Code, my goal is to make these solutions as readable as possible. These are the ones I have completed so far.

1. Multiples of 3 or 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.

My Solution

total = 0
for potential_multiple in range(1000):
  if(potential_multiple % 3 == 0 or potential_multiple % 5 == 0):
    total += potential_multiple
print(total)

Browser Executable Version

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.

My Solution

first_term = 1
second_term = 2
total = 2

while(first_term < 4e6 and second_term < 4e6):
  first_term += second_term
  if first_term % 2 == 0:
    total += first_term
  second_term += first_term
  if second_term % 2 == 0:
    total += second_term
print(total)

Browser Executable Version

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 ?

My Solution

import math

def largest_prime_factor(factorable_number):
  prime_list = find_smaller_primes(factorable_number)
  for prime in reversed(prime_list):
    if factorable_number % prime == 0:
      return prime
  return "no prime factors."

def find_smaller_primes(upper_bound):
  primes = []
  while upper_bound % 2 == 0:
    primes.append(2)
    upper_bound //= 2

  divisor = 3
  while divisor * divisor <= upper_bound:
    if upper_bound % divisor == 0:
      primes.append(divisor)
      upper_bound //= divisor
    else:
      divisor += 2

  if upper_bound != 1: primes.append(upper_bound)
  return primes
  
print(largest_prime_factor(600851475143))

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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.

My Solution

def largest_palindrome_product(number_of_digits):
  upper_bound = ""
  largest_palindrome = 0
  for each in range(0, number_of_digits):
    upper_bound += "9"
  upper_bound = int(upper_bound)
  while upper_bound ** 2 > largest_palindrome:
    for second_factor in reversed(range(1, upper_bound+1)):
      if is_palindrome(second_factor * upper_bound):
        largest_palindrome = max(largest_palindrome, second_factor * upper_bound)
        break
    upper_bound -= 1
  return largest_palindrome

def is_palindrome(potential_palindrome):
  letters = str(potential_palindrome)
  check_from_beginning = 0
  for check_from_end in reversed(letters):
    if check_from_end != letters[check_from_beginning]:
      return False
    check_from_beginning += 1
  return True

print(largest_palindrome_product(3))

Browser Executable Version

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?

My Solution

def least_common_multiple(upper_bound):
  possible_factors_remaining = [each for each in range(2, upper_bound+1)]
  primes_and_powers = {}
  
  while possible_factors_remaining:
    prime_factor = possible_factors_remaining[0]
    primes_and_powers[prime_factor] = 1
    primes_and_powers = find_nonprime_factors(primes_and_powers, prime_factor, upper_bound)
    possible_factors_remaining = remove_multiples(possible_factors_remaining, prime_factor)
  
  total = 1
  for factor, exponent in primes_and_powers.items():
    total *= factor ** exponent
  return total

def find_nonprime_factors(primes_and_powers, prime_factor, upper_bound):
    power = 2
    while prime_factor ** power < upper_bound:
      primes_and_powers[prime_factor] = primes_and_powers.get(prime_factor) + 1
      power += 1
    return primes_and_powers
  
def remove_multiples(possible_factors_remaining, prime_factor):
    for multiple in possible_factors_remaining:
      if multiple % prime_factor == 0:
        possible_factors_remaining.remove(multiple)
    return possible_factors_remaining
  

print(least_common_multiple(20))

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6. Sum Square Difference

The sum of the squares of the first ten natural numbers is,

The square of the sum of the first ten natural numbers is,

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is .

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

My Solution

def square_of_sum(upper_bound):
    sum_to_be_squared = 0
    for each in range(1, upper_bound+1):
        sum_to_be_squared += each
    return sum_to_be_squared ** 2
	
def sum_of_squares(upper_bound):
    sum_of_squares = 0
    for each in range(1, upper_bound+1):
        sum_of_squares += each ** 2
    return sum_of_squares

print(square_of_sum(100) - sum_of_squares(100))

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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?

My Solution

import math


def is_prime(potential_prime):
  if (potential_prime == 1): return False
  if (potential_prime < 4): return True
  if (potential_prime % 2 == 0): return False
  if (potential_prime < 9): return True
  if (potential_prime % 3 == 0): return False
  max_prime_factor = math.floor(math.sqrt(potential_prime))
  prime_factor = 5
  while prime_factor <= max_prime_factor:
    if potential_prime % prime_factor == 0: return False
    if potential_prime % (prime_factor + 2) == 0: return False
    prime_factor += 6
  return True


def nth_prime(n):
  count = 1
  candidate = 1
  while (count < n):
    candidate = candidate + 2
    if (is_prime(candidate)): count += 1
  return candidate


print(nth_prime(10001))

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8. Largest product in a series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 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?

My Solution

import math

def product_finder(sequence_with_newlines, digits):
  if len(sequence_with_newlines) < digits: raise ValueError("Sequence is too short.")
  sequence = [int(every_number) for every_number in sequence_with_newlines if every_number != "\n"] 
  candidates = []
  for candidate_initialization in range(0, digits):
    candidates.append(sequence.pop(0))
  product = 1
  while(sequence):
    candidates.remove(candidates[0])
    candidates.append(sequence.pop(0))
    product = max(product, math.prod(candidates))
  return product

sequence = '''73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'''

print(product_finder(sequence, 13))

Browser Executable Version

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, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

My Solution

import math

def pythagorean_triple_product(total):
  for m in reversed(range(1, int(math.sqrt(total))+1)):
    n = (total - 2*m**2) / (2*m)
    a,b,c = euclid_formula(m,n)
    if (a>0) & (b>0) & (c>0): return a*b*c
  return 0

def euclid_formula(m, n):
    if not ((m > n) & (n > 0)): return 0,0,0
    if int(n) != float(n): return 0,0,0
    if not ((m % 2 == 0) ^ (n % 2 == 0)): return 0,0,0
    
    a = m**2 - n**2
    b = 2*m*n
    c = m**2 + n**2
    return a,b,c

print(pythagorean_triple_product(1000))

Browser Executable Version

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.

My Solution

import math


def prime_sum(upper_bound):
  is_prime = [True] * upper_bound
  is_prime[0] = False
  is_prime[1] = False
  max_prime_factor = math.floor(math.sqrt(upper_bound))

  for prime in range(2, max_prime_factor + 1):
    multiple = prime * 2
    while multiple < upper_bound:
      is_prime[multiple] = False
      multiple += prime

  prime = []

  for confirmed_prime in range(upper_bound):
    if is_prime[confirmed_prime] == True:
      prime.append(confirmed_prime)

  return sum(prime)


print(prime_sum(2000000))

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11. Largest Product in a Grid

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

My Solution

#still working on this one
#It should be the sequence reader from #8 and then feed in all the directions.
#Plus some more changes since the numbers in the sequence are 2-digit.
#Not bad at all.
import math

def main(grid, digits):
	array = grid.split("\n")
	print(array)
	print(len(array)) #confirmed 20
	#convert to 2D array
	#split into sequences, send to product_finder
	#return largest product


def product_finder(sequence_with_newlines, digits):
  if len(sequence_with_newlines) < digits: raise ValueError("Sequence is too short.")
  sequence = [int(every_number) for every_number in sequence_with_newlines if every_number != "\n"] 
  candidates = []
  for candidate_initialization in range(0, digits):
    candidates.append(sequence.pop(0))
  product = 1
  while(sequence):
    candidates.remove(candidates[0])
    candidates.append(sequence.pop(0))
    product = max(product, math.prod(candidates))
  return product

grid = '''08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48'''

print(main(grid, 4))

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