Adam Kilgour Portfolio

Project Euler

Task 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 $1000$.

top_num = 1000
nums = set(range(0,top_num, 3)) | set(range(0,top_num, 5))
print(sum(nums))
# Result is 233168

$$ 3+5+6+9+10 \cdots+996=233168 $$

Task 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: $$ 1,2,3,5,8,13,21,34,55,89 \dots $$ By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

n  = 4000000
vals = [1,2]
while True:
  if vals[-1] + vals[-2] < n:
    vals.append(vals[-1] + vals[-2])
  else:
    break

evenVals = filter(lambda x: x % 2 == 0, vals)
output = sum(list(evenVals))
print(output)
# output is 4613732

Task 3 Largest Prime Factor

The prime factors of $13195$ are $5,7,13,29$ what is the largest prime factor of $600851475143$

num = 600851475143
vals = set(range(3, int(num ** 0.5), 2))
primes = [1]

while len(vals) > 1 :
primes.append(min(vals))
tempFilter = set(range(min(vals), int(num ** 0.5), min(vals)))
vals = vals - tempFilter

possibleFactors = primes
while True:
if num % max(possibleFactors) == 0:
    print(max(possibleFactors))
    break
else:
    possibleFactors.remove(max(possibleFactors))
  #6857

Task 4 palindrome number

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

def isPalindrome(num):
    num = [int(digit) for digit in str(num)]
    return num == num[::-1]
  
  a = 999
  b = 999
  while True:
      if isPalindrome(a * b):
          print(a * b)
          print(a)
          print(b)
          break
      else:
          if a == 100:
              a = 999
              b -= 1
          else:
              a -= 1
  # 583 * 995 = 580085

Task 5 smallest multiple

divisor = 2
  number = 1
  while divisor < 20:
      number +=1
      divisor = 2
      while True:
          if number % divisor != 0:
              break
          else:
              divisor +=1
  
  print(f"The number {number} is divisable by all numbers up to {divisor - 1}")
  # 232792560

sum square difference

   num = 100
  SquareOfSum = sum(range(num + 1)) ** 2
  sumOfSquares = sum([i ** 2 for i in range(num + 1)])
  print(f"{SquareOfSum} - {sumOfSquares} = {SquareOfSum - sumOfSquares}" )
  # 25502500 - 338350 = 25164150