Python/project_euler/problem_123/sol1.py
Ravi Kandasamy Sundaram c8db6a208b
Add solution for Project Euler 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.

Reference: https://projecteuler.net/problem=123

reference: 

Co-authored-by: Ravi Kandasamy Sundaram <rkandasamysundaram@luxoft.com>
2020-11-06 22:25:02 +05:30

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"""
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 (pn1)^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 typing import Dict, 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())