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* Solution to Problem 145 of Project Euler * Provided more descriptive filename * Update sol1.py Co-authored-by: John Law <johnlaw.po@gmail.com>
58 lines
1.6 KiB
Python
58 lines
1.6 KiB
Python
"""
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Project Euler problem 145: https://projecteuler.net/problem=145
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Author: Vineet Rao
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Problem statement:
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Some positive integers n have the property that the sum [ n + reverse(n) ]
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consists entirely of odd (decimal) digits.
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For instance, 36 + 63 = 99 and 409 + 904 = 1313.
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We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
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Leading zeroes are not allowed in either n or reverse(n).
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There are 120 reversible numbers below one-thousand.
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How many reversible numbers are there below one-billion (10^9)?
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"""
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def odd_digits(num: int) -> bool:
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"""
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Check if the number passed as argument has only odd digits.
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>>> odd_digits(123)
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False
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>>> odd_digits(135797531)
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True
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"""
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num_str = str(num)
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for i in ["0", "2", "4", "6", "8"]:
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if i in num_str:
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return False
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return True
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def solution(max_num: int = 1_000_000_000) -> int:
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"""
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To evaluate the solution, use solution()
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>>> solution(1000)
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120
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>>> solution(1_000_000)
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18720
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>>> solution(10_000_000)
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68720
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"""
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result = 0
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# All single digit numbers reverse to themselves, so their sums are even
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# Therefore at least one digit in their sum is even
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# Last digit cannot be 0, else it causes leading zeros in reverse
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for num in range(11, max_num):
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if num % 10 == 0:
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continue
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num_sum = num + int(str(num)[::-1])
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num_is_reversible = odd_digits(num_sum)
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result += 1 if num_is_reversible else 0
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return result
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if __name__ == "__main__":
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print(f"{solution() = }")
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