Add solution to Problem 145 of Project Euler (#5464)

* Solution to Problem 145 of Project Euler

* Provided more descriptive filename

* Update sol1.py

Co-authored-by: John Law <[email protected]>

Author: VinWare

project_euler/problem_145/__init__.py

@@ -0,0 +1,57 @@
+"""
+Project Euler problem 145: https://projecteuler.net/problem=145
+Author: Vineet Rao
+Problem statement:
+
+Some positive integers n have the property that the sum [ n + reverse(n) ]
+consists entirely of odd (decimal) digits.
+For instance, 36 + 63 = 99 and 409 + 904 = 1313.
+We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
+Leading zeroes are not allowed in either n or reverse(n).
+
+There are 120 reversible numbers below one-thousand.
+
+How many reversible numbers are there below one-billion (10^9)?
+"""
+
+
+def odd_digits(num: int) -> bool:
+    """
+    Check if the number passed as argument has only odd digits.
+    >>> odd_digits(123)
+    False
+    >>> odd_digits(135797531)
+    True
+    """
+    num_str = str(num)
+    for i in ["0", "2", "4", "6", "8"]:
+        if i in num_str:
+            return False
+    return True
+
+
+def solution(max_num: int = 1_000_000_000) -> int:
+    """
+    To evaluate the solution, use solution()
+    >>> solution(1000)
+    120
+    >>> solution(1_000_000)
+    18720
+    >>> solution(10_000_000)
+    68720
+    """
+    result = 0
+    # All single digit numbers reverse to themselves, so their sums are even
+    # Therefore at least one digit in their sum is even
+    # Last digit cannot be 0, else it causes leading zeros in reverse
+    for num in range(11, max_num):
+        if num % 10 == 0:
+            continue
+        num_sum = num + int(str(num)[::-1])
+        num_is_reversible = odd_digits(num_sum)
+        result += 1 if num_is_reversible else 0
+    return result
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")