mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-07 14:19:31 +00:00
101 lines
2.3 KiB
Python
101 lines
2.3 KiB
Python
"""
|
||
Problem 123: https://projecteuler.net/problem=123
|
||
|
||
Name: Prime square remainders
|
||
|
||
Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and
|
||
let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.
|
||
|
||
For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
|
||
The least value of n for which the remainder first exceeds 10^9 is 7037.
|
||
|
||
Find the least value of n for which the remainder first exceeds 10^10.
|
||
|
||
|
||
Solution:
|
||
|
||
n=1: (p-1) + (p+1) = 2p
|
||
n=2: (p-1)^2 + (p+1)^2
|
||
= p^2 + 1 - 2p + p^2 + 1 + 2p (Using (p+b)^2 = (p^2 + b^2 + 2pb),
|
||
(p-b)^2 = (p^2 + b^2 - 2pb) and b = 1)
|
||
= 2p^2 + 2
|
||
n=3: (p-1)^3 + (p+1)^3 (Similarly using (p+b)^3 & (p-b)^3 formula and so on)
|
||
= 2p^3 + 6p
|
||
n=4: 2p^4 + 12p^2 + 2
|
||
n=5: 2p^5 + 20p^3 + 10p
|
||
|
||
As you could see, when the expression is divided by p^2.
|
||
Except for the last term, the rest will result in the remainder 0.
|
||
|
||
n=1: 2p
|
||
n=2: 2
|
||
n=3: 6p
|
||
n=4: 2
|
||
n=5: 10p
|
||
|
||
So it could be simplified as,
|
||
r = 2pn when n is odd
|
||
r = 2 when n is even.
|
||
"""
|
||
from __future__ import annotations
|
||
|
||
from collections.abc import Generator
|
||
|
||
|
||
def sieve() -> Generator[int, None, None]:
|
||
"""
|
||
Returns a prime number generator using sieve method.
|
||
>>> type(sieve())
|
||
<class 'generator'>
|
||
>>> primes = sieve()
|
||
>>> next(primes)
|
||
2
|
||
>>> next(primes)
|
||
3
|
||
>>> next(primes)
|
||
5
|
||
>>> next(primes)
|
||
7
|
||
>>> next(primes)
|
||
11
|
||
>>> next(primes)
|
||
13
|
||
"""
|
||
factor_map: dict[int, int] = {}
|
||
prime = 2
|
||
while True:
|
||
factor = factor_map.pop(prime, None)
|
||
if factor:
|
||
x = factor + prime
|
||
while x in factor_map:
|
||
x += factor
|
||
factor_map[x] = factor
|
||
else:
|
||
factor_map[prime * prime] = prime
|
||
yield prime
|
||
prime += 1
|
||
|
||
|
||
def solution(limit: float = 1e10) -> int:
|
||
"""
|
||
Returns the least value of n for which the remainder first exceeds 10^10.
|
||
>>> solution(1e8)
|
||
2371
|
||
>>> solution(1e9)
|
||
7037
|
||
"""
|
||
primes = sieve()
|
||
|
||
n = 1
|
||
while True:
|
||
prime = next(primes)
|
||
if (2 * prime * n) > limit:
|
||
return n
|
||
# Ignore the next prime as the reminder will be 2.
|
||
next(primes)
|
||
n += 2
|
||
|
||
|
||
if __name__ == "__main__":
|
||
print(solution())
|