def test(number, end=2):
number_digits = set(num2digits(number))
for x in range(2, end + 1):
multiplied = number * x
multiplied_digits = set(num2digits(multiplied))
if not number_digits == multiplied_digits:
return False
return True
def is_palindrome(number):
digits = num2digits(number)
size = len(digits)
for i in range(size // 2):
if digits[i] != digits[size - i - 1]:
return False
return True
def pandigital(num, digitsNumber=9):
digits = []
multiplier = 1
while len(digits) < digitsNumber:
multiplied = num * multiplier
digits.extend(num2digits(multiplied))
multiplier += 1
return digits
def permute(seed):
seedDigits = num2digits(seed)
legalDigits = list(set(range(1, 10)).difference(seedDigits))
legalDigits.sort(reverse=True)
for i in legalDigits:
yield seed * 10 + i
'''
A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large:
one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
Considering natural numbers of the form, ab, where a, b < 100, what is the maximum digital sum?
'''
from src.utils.digits import num2digits
max_sum = -1
for i in reversed(range(100)):
if i % 10 == 0:
continue
for j in reversed(range(100)):
digits_sum = sum(map(int, num2digits(i**j)))
if digits_sum > max_sum:
max_sum = digits_sum
print(max_sum)
def reverse_add(number):
digits = num2digits(number)
digits.reverse()
reversed_number = digits2num(digits)
return number + reversed_number
def digit_gen():
current = 1
while True:
for digit in num2digits(current):
yield digit
current += 1