Project Euler was started by Colin Hughes (a.k.a. euler) in October 2001 as a sub-section on mathschallenge.net. Who could have known how popular these types of problems would turn out to be? Since then the membership has continued to grow and Project Euler moved to its own domain in 2006.
“Project Euler exists to encourage, challenge, and develop the skills and enjoyment of anyone with an interest in the fascinating world of mathematics.”
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.
First, I created a cached function in order to store all the calculated values, then I run a while loop to sum the values.
fib_cache = {}
def fib(n):
if n in fib_cache: return fib_cache[n]
fib_cache[n] = n if n < 3 else fib(n - 1) + fib(n - 2)
return fib_cache[n]
i = 1
sum = res = 0
while res <= 4_000_000:
res = fib(i)
if res % 2 == 0: sum += res
i = i + 1
print(sum) # 4613732
Another simpler approach is using python swap constructor:
res = 0
a = b = 1
while b <= 4_000_000:
a, b = b, a + b
if b % 2 == 0: res += b
print(res) # 4613732
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.
def maxPalindrome():
max = 0
for i in range(1, 1000):
for j in range(1, 1000):
res = i * j
if str(res) == str(res)[::-1] and res > max: max = res
return max
print(maxPalindrome())
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?
def divisible(num):
divisible = False
n = num
while divisible == False:
for divisor in range(1, num + 1):
if n % divisor != 0:
n += 1
break
if divisor == num:
divisible = True
return n
print(divisible(20)) #232792560
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.
def triplet(limit):
for a in range(1,limit + 1):
for b in range(1, limit + 1 - a):
c = 1000 - a - b
if a*a + b*b == c*c:
return a*b*c
print(triplet(1000)) #31875000
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?
from functools import reduce
# (used regex on sublime to set the num in one line)
num = str(7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450)
chunk = 13
# create a list of all possible 13-digits long chunks
nums = [num[i: i + chunk] for i in range(0, len(num) - chunk + 1, 1)]
# create a list "factor", with lists containing one element per digit
factors = [[int(n) for n in num] for num in nums]
# apply the product of each factor within the lists, separately
products = [reduce(lambda x, y: x*y, factor) for factor in factors]
# max product found
print(max(products)) # 23514624000
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
import math as mth
def isPrime(num):
if num == 1: return False
if num == 2: return True
if not num & 1: return False
limit = int(mth.sqrt(num)) + 1
for div in range(3, limit, 2):
if num % div == 0: return False
return True
sum = 0
for num in range(2, 2_000_000):
if isPrime(num): sum += num
print(sum) # 142913828922