Added solution to Project Euler 69

project_euler/problem_69/sol1.py

@@ -0,0 +1,66 @@
+"""
+Totient maximum
+Problem 69: https://projecteuler.net/problem=69
+
+Euler's Totient function, φ(n) [sometimes called the phi function],
+is used to determine the number of numbers less than n which are relatively prime to n.
+For example, as 1, 2, 4, 5, 7, and 8,
+are all less than nine and relatively prime to nine, φ(9)=6.
+
+n	Relatively Prime	φ(n)	n/φ(n)
+2	1	                1	    2
+3	1,2	                2	    1.5
+4	1,3	                2	    2
+5	1,2,3,4	            4	    1.25
+6	1,5		            2	    3
+7	1,2,3,4,5,6	        6	    1.1666...
+8	1,3,5,7		        4	    2
+9	1,2,4,5,7,8	        6	    1.5
+10	1,3,7,9	            4	    2.5
+
+It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
+
+Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
+"""
+
+
+def solution(n: int = 10 ** 6) -> int:
+    """
+    Returns solution to problem.
+    Algorithm:
+    1. Precompute φ(k) for all natural k, k <= n using product formula (wikilink below)
+    https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler's_product_formula
+
+    2. Find k/φ(k) for all k ≤ n and return the k that attains maximum
+
+    >>> solution(10)
+    6
+
+    >>> solution(100)
+    30
+
+    >>> solution(9973)
+    2310
+
+    """
+
+    if n <= 0:
+        raise ValueError("Please enter an integer greater than 0")
+
+    phi = list(range(n + 1))
+    for number in range(2, n + 1):
+        if phi[number] == number:
+            phi[number] -= 1
+            for multiple in range(number * 2, n + 1, number):
+                phi[multiple] = (phi[multiple] // number) * (number - 1)
+
+    answer = 1
+    for number in range(1, n + 1):
+        if (answer / phi[answer]) < (number / phi[number]):
+            answer = number
+
+    return answer
+
+
+if __name__ == "__main__":
+    print(solution())