project_euler.problem_123.sol1

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.

Functions

sieve(→ collections.abc.Generator[int])	Returns a prime number generator using sieve method.
solution(→ int)	Returns the least value of n for which the remainder first exceeds 10^10.

Module Contents

project_euler.problem_123.sol1.sieve() → collections.abc.Generator[int]

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

project_euler.problem_123.sol1.solution(limit: float = 10000000000.0) → int

Returns the least value of n for which the remainder first exceeds 10^10. >>> solution(1e8) 2371 >>> solution(1e9) 7037