Python/project_euler/problem_069/sol1.py
Dhruv 44254cf112
Rename Project Euler directories and other dependent changes (#3300)
* Rename all Project Euler directories:

Reason:
The change was done to maintain consistency throughout the directory
and to keep all directories in sorted order.

Due to the above change, some config files had to be modified:
'problem_22` -> `problem_022`

* Update scripts to pad zeroes in PE directories
2020-10-15 12:43:28 +05:30

67 lines
1.7 KiB
Python

"""
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())