``````

In [1]:

import notebook_importer
from shared_functions import fib_gen, factorization, primesfrom2to
from gmpy2 import is_prime
import itertools
import numpy as np
import os

``````
``````

importing notebook from shared_functions.ipynb

``````

# 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.

``````

In [2]:

sum(i for i in range(1000) if i%3 == 0 or i%5 == 0)

``````
``````

Out[2]:

233168

``````

# 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.

``````

In [17]:

sum(i for i in fib_gen(4000000) if i%2 == 0)

``````
``````

Out[17]:

4613732

``````

# 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

``````

In [4]:

def find_large_prime_factor(n):
for i in itertools.count(start=2):
if n % i == 0 and is_prime(n // i):
return n // i

find_large_prime_factor(600851475143)

``````
``````

Out[4]:

6857

``````

# 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.

``````

In [5]:

def is_palindrome_num(a):
return str(a) == str(a)[::-1]

def find_large_palindrome_prod():
max_r = 0
for a in reversed(range(100, 999)):
for b in reversed(range(100, a+1)):
r = a*b
if is_palindrome_num(r) and r > max_r:
max_r = r
return max_r

find_large_palindrome_prod()

``````
``````

Out[5]:

906609

``````

# 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?

``````

In [6]:

def find_smallest_evenly_divisible(n):
result = []
for i in range(2,n+1):
new_r = list(result)
for f in factorization(i):
try:
del new_r[new_r.index(f)]
except ValueError:
result.append(f)
return np.prod(result)

find_smallest_evenly_divisible(20)

``````
``````

Out[6]:

232792560

``````

# 6: Sum square difference

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 = 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 [7]:

in_arr = np.arange(1,101)
sum_of_squares = np.power(np.sum(in_arr),2)
square_of_sum = np.sum(np.apply_along_axis(np.power, 0, in_arr, 2))
sum_of_squares - square_of_sum

``````
``````

Out[7]:

25164150

``````

# 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?

``````

In [8]:

for i,p in enumerate(primesfrom2to(1000000), start=1):
if i == 10001:
break
p

``````
``````

Out[8]:

104743

``````

# 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?

``````

In [9]:

num_in = """73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
7163626956188267042825248360082325753042075296345"""

length = 13
num_in = num_in.replace(os.linesep,"")

max_prod = 0
max_seg = ""
for i in range(len(num_in)-length):
prod = np.prod(list(map(int,num_in[i:i+13])))
if prod > max_prod:
max_prod = prod
max_seg = num_in[i:i+13]

max_prod

``````
``````

Out[9]:

23514624000

``````

# 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, \$3^2 + 4^2 = 9 + 16 = 25 = 5^2\$

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

``````

In [10]:

perfect_squares = np.apply_along_axis(np.power,0,np.arange(1,1000),2)

for a in perfect_squares:
b_indx = np.searchsorted(perfect_squares, a)
for b in perfect_squares[b_indx:]:
c = a + b
if c in perfect_squares and np.sqrt(a)+np.sqrt(b)+np.sqrt(c) == 1000:
result = int(np.sqrt(a)*np.sqrt(b)*np.sqrt(c))
break
result

``````
``````

Out[10]:

31875000

``````

# 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.

``````

In [11]:

np.sum(primesfrom2to(2000000))

``````
``````

Out[11]:

142913828922

``````