Euler 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.
My Solution
import numpy as np
def sum_of_multiples(multiples=[3,5],upper_lim=1000):
# count multiples
first = sum(np.arange(0,upper_lim,multiples[0]))
second = sum(np.arange(0,upper_lim,multiples[1]))
# remove common multiples so don't double count
common_multiples = multiples[0]*multiples[1]
common = sum(np.arange(0,upper_lim,common_multiples))
sum_ = first + second - common
return sum_
sum_of_multiples()
Answer: 233,168
Euler 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.
My Solution
def sum_even_fib(upper_lim=4000000):
k0 = 1
k1 = 2
k = 3
total = 2
while (k<=upper_lim):
k0 = k1
k1 = k
k = k0 + k1
if k%2==0:
total+=k
return total
sum_even_fib()
Answer: 4,613,732
Euler Problem 3
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
My Solution
import numpy as np
def largest_prime_factor(n):
# Function to check if number is prime
# checks numbers from 2 to sqrt of n to save time
def check_prime(n):
for a in range(2,int(n**0.5)+1):
if n % a == 0:
return False
return True
# check factors and update largest prime found
largest_prime = np.nan
for a in range(2,int(np.sqrt(n))):
if (n % a == 0) & (check_prime(a)):
largest_prime = a
return largest_prime
largest_prime_factor(600851475143)
Answer: 6,857
Euler 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 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
My Solution
def is_palindrome(n):
return str(n) == str(n)[::-1]
def largest_3_digit():
largest = 0
# Look at 3 digit combos
for a in reversed(range(100,999)):
for b in reversed(range(100,999)):
# check if palindrome and if larger than current value
if (is_palindrome(a*b)) & (a*b>largest): largest = a*b
return largest
largest_3_digit()
Answer: 906,609
Euler 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.
My Solution
import numpy as np
def sum_of_sq_and_sq_of_sum():
one_hund = np.arange(1,101)
sum_of_sq = sum(one_hund**2)
sq_of_sum = sum(one_hund)**2
return sq_of_sum-sum_of_sq
sum_of_sq_and_sq_of_sum()
Answer: 25,164,150
Euler 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?
My Solution
# check if number is prime
def check_prime(n):
for a in range(2,int(n**0.5)+1):
if n % a == 0:
return False
return True
# Start with initial primes and add to list if prime
def get_primes(n):
primes =[2, 3, 5]
num = 6
while len(primes)<n:
if check_prime(num) : primes.append(num)
num+=1
return primes
primes = get_primes(10001)
primes[-1]
Answer: 104,743
Euler Problem 8
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
Problem Info
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
num = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
def find_largest_product(num,n):
str_num = str(num)
largest_product = 0
for j in range(0,len(str_num)-n):
product_str = str_num[j:j+n]
product = 1
for char in product_str:
product = product * int(char)
if product > largest_product: largest_product = product
return largest_product
find_largest_product(num,13)
''' Other solution found online
max_prod, adj = 0, 13
for i in range(len(digs)-adj):
max_prod = max(max_prod, math.prod([int(d) for d in digs[i:i+adj]]))
print(max_prod)
'''
Answer: 23,514,624,000
Euler 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.
My Solution
def sum_primes(upper_lim=2000000):
arr = range(2,upper_lim)
sum = 0
def check_prime(n):
for a in range(2,int(n**0.5)+1):
if n % a == 0:
return False
return True
for a in arr:
if check_prime(a):
sum+=a
return sum
Answer: 142,913,828,922