In [ ]:
import math
from operator import mul
from functools import reduce
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.
In [1]:
#Set up the sum of multiples and reset to zero
multiples_sum = 0
#Create the function that divides all the numbers from 1-1000 by 3 or 5 and add them
for i in range(1, 1000):
if (i % 3 == 0 or i % 5 == 0):
multiples_sum = multiples_sum + i
#Print results
print (multiples_sum)
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.
In [5]:
#Create empty dictionary to populate wth the sequence of fibonacci terms
sum_even_valued = {}
#Define fibonnaci function that will find the terms of the fibonacci series
def fibonacci(n):
sum_even_valued[n] = sum_even_valued.get(n, 0) or (n <= 1 and 1 or fiba(n-1) + fiba(n-2))
return sum_even_valued[n]
In [6]:
#Reset n and x to find the fibonaccy terms and add them if they are smaller than 4000000
n = 0
x = 0
#Add the fibonaccy terms that are lower than 4000000
while fibonacci(x) <= 4000000:
if not fibonacci(x) % 2: n = n + fibonacci(x)
x=x+1
#Print results
print(n)
Problem 3
The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?
In [10]:
#Set the max number of he range of numbers that contain the prie numbers
num=600851475143
#Define a break for the series of numbers to be found starting by two
#These are the divisors
i=2
#Define the function that checks if the module of the number is zero
#Check if the number is
for k in range(0,num):
if i >= num:
break
elif num % i == 0: # Check if the number is evenly divisible by i
num = num / i
else:
i= i + 1
#Print the result
print ("biggest prime number is: "+str(num))
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.
In [11]:
#Start the number
n = 0
#check all the multiplications made of numbers of three digits and stop when read in both directions is equal
for a in range(999, 100, -1):
for b in range(a, 100, -1):
x = a * b
if x > n:
s = str(a * b)
if s == s[::-1]:
n = a * b
#Print results
print(n)
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?
In [46]:
#Define the functions that finds the samallest number
def smallest(n):
for i in range(n, factorial(n) + 1, n):
if multiple(i, n):
return i
return -1
#Define the number that is a multiple of the range of numbers
def multiple(x, n):
for i in range(1, n):
if x % i != 0:
return False
return True
#Define the factoral function
def factorial(n):
if n > 1: return n * factorial(n - 1)
elif n >= 0: return 1
else: return -1
print (smallest(10))
print (smallest(20))
Problem 6 The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385 The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 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.
In [16]:
#Define the range of numbers we are going to check
r = range(1, 101)
a = sum(r)
#Multipy the sum of the number and subtract the suqae of the sum of the numbers
print (a * a - sum(i*i for i in r))
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?
In [42]:
#Define the function that finds the primer numbers
def primes():
D = {}
q = 2
while 1:
if q not in D:
yield q
D[q*q] = [q]
else:
for p in D[q]:
D.setdefault(p+q,[]).append(p)
del D[q]
q += 1
#Define the function that counts the position of the number
def nth_prime(n):
for i, prime in enumerate(primes()):
if i == n - 1:
return prime
print(nth_prime(10001))
Problem 8 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?
In [39]:
st = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'
In [41]:
size = 13
max(reduce(mul,map(int,st[i:i + size])) for i in range(len(st) - size))
Out[41]:
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.
In [31]:
#check all the numbers from 1 to 1000 that fulfill the a+b+c = 1000 condition
#if the lowest number fulfills the condition check the product of the squares
for a in range(1, 1000):
for b in range(a, 1000):
c = 1000 - a - b
if c > 0:
if c*c == a*a + b*b:
print (a*b*c)
break
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.
In [48]:
#Define the number that is prime
def prime(num):
if num > 2 and num % 2 == 0:
return False
else:
for i in range(3, int(math.sqrt(num)) + 1, 2):
if num % i == 0:
return False
return True
#Define a function that is adding up to the limit
def adding_up(limit):
sum = 0
for i in range(2, limit):
if prime(i):
sum += i
return sum
In [49]:
print(adding_up(2000000))