# Problem 53
from math_stuff import factorial
total = 0
for n in range(23, 101):
for r in range(1, n + 1):
nr = factorial(n) / (factorial(r) * factorial(n - r))
if nr > 1_000_000:
total += 1
print(total)
# Problem 20
#
# n! means n × (n − 1) × ... × 3 × 2 × 1
#
# For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
# and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
#
# Find the sum of the digits in the number 100!
from math_stuff import factorial
fact_100 = factorial(100)
my_sum = 0
for d in str(fact_100):
my_sum += int(d)
print(my_sum)
# Problem 34
# 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
# Find the sum of all numbers which are equal to the sum of the factorial of
# their digits.
# Note: as 1! = 1 and 2! = 2 are not sums they are not included.
# 7*9! is a 7 digit number. 8*9! is also a 7 digit number so no answer will be
# greater than 7 digits.
from math_stuff import factorial
def sum_of_fact_of_dig(number):
""" Returns the sum of the factorial of numbers digits """
ret_sum = 0
for d in str(number):
ret_sum += factorial(int(d))
return ret_sum
upper_bound = 7 * factorial(9)
all_sums = 0
for i in range(10,upper_bound + 1):
if i == sum_of_fact_of_dig(i):
all_sums += i
print(all_sums)
def sum_of_fact_of_dig(number):
""" Returns the sum of the factorial of numbers digits """
ret_sum = 0
for d in str(number):
ret_sum += factorial(int(d))
return ret_sum
# Problem 24 v2
#
# Instead of computing every permutation, this will only compute the
# permutation at the requested position
from math_stuff import factorial
import time
start_time = time.time()
numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
numbers.sort
position = 1_000_000
if position > factorial(len(numbers)):
print('nope :(')
exit()
mutation = ''
while len(numbers) > 1:
j = position % factorial(len(numbers))
k = 0
index = 0
while k < j:
index += 1
k = factorial(len(numbers) - 1) * index
mutation += str(numbers.pop(index - 1))
mutation += str(numbers.pop())
print(mutation)