feat: 27 - quadratic primes

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newt 2024-10-09 18:10:12 +01:00
parent 525ba998df
commit 19742c35b0
3 changed files with 78 additions and 6 deletions

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@ -16,7 +16,7 @@ All of the solutions here are written in rust. [main.rs](src/main.rs) is the hom
## Challenge Completion
### <!-- completed -->21<!-- completed --> out of 100 public challenges completed.
### <!-- completed -->22<!-- completed --> out of 100 public challenges completed.
- [x] 1 - [Multiples of 3 or 5](src/bin/1.rs)
- [x] 2 - [Even Fibonacci numbers](src/bin/2.rs)
@ -44,7 +44,7 @@ All of the solutions here are written in rust. [main.rs](src/main.rs) is the hom
- [ ] 24 - Lexicographic permutations
- [ ] 25 - 1000-digit Fibonacci number
- [ ] 26 - Reciprocal cycles
- [ ] 27 - Quadratic primes
- [x] 27 - [Quadratic primes](src/bin/27.rs)
- [ ] 28 - Number spiral diagonals
- [ ] 29 - Distinct powers
- [ ] 30 - Digit fifth powers
@ -119,4 +119,4 @@ All of the solutions here are written in rust. [main.rs](src/main.rs) is the hom
- [ ] 99 - Largest exponential
- [ ] 100 - Arranged probability
<sub>Check out Project Euler <a href="https://projecteuler.net">here</a>.</sub>
<sub>Check out Project Euler <a href="https://projecteuler.net">here</a>.</sub>

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src/bin/27.rs Normal file
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/*
Problem 27 - Quadratic primes
Euler discovered the remarkable quadratic formula:
$n^2 + n + 41$
It turns out that the formula will produce 40 primes for the consecutive integer values $0 \le n \le 39$. However, when $n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41$ is divisible by 41, and certainly when $n = 41, 41^2 + 41 + 41$ is clearly divisible by 41.
The incredible formula $n^2 - 79n + 1601$ was discovered, which produces 80 primes for the consecutive values $0 \le n \le 79$. The product of the coefficients, 79 and 1601, is 126479.
Considering quadratics of the form:
$n^2 + an + b$, where $|a| < 1000$ and $|b| \le 1000$ where $|n|$ is the modulus/absolute value of $n$ e.g. $|11| = 11$ and $|-4| = 4$
Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.
*/
use std::ops::Mul;
const PRIMES: [i32; 168] = [
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193,
197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307,
311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421,
431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547,
557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929,
937, 941, 947, 953, 967, 971, 977, 983, 991, 997,
];
fn is_prime(n: i32) -> bool {
if n <= 1 {
return false;
}
for a in 2..n {
if n % a == 0 {
return false;
}
}
true
}
fn consecutive_primes(a: i32, b: i32) -> i32 {
let mut n = 0;
loop {
let t = (n + a).mul(n) + b;
if !is_prime(t) {
return n;
}
n += 1;
}
}
pub fn main() {
let mut max_c = 0;
let mut max_ab = 0;
for a in (-999..=1001i32).step_by(2) {
for i in 0..PRIMES.len() {
let b = PRIMES[i];
let c = consecutive_primes(a - (if i == 0 { 1 } else { 0 }), b);
if c > max_c {
max_c = c;
max_ab = a * b;
}
}
}
println!("{:?}", max_ab);
}

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@ -52,8 +52,8 @@ mod twenty_two;
// mod twenty_five;
// #[path = "../bin/26.rs"]
// mod twenty_six;
// #[path = "../bin/27.rs"]
// mod twenty_seven;
#[path = "../bin/27.rs"]
mod twenty_seven;
// #[path = "../bin/28.rs"]
// mod twenty_eight;
// #[path = "../bin/29.rs"]
@ -240,7 +240,7 @@ pub async fn execute(
// 24 => twenty_four::main(),
// 25 => twenty_five::main(),
// 26 => twenty_six::main(),
// 27 => twenty_seven::main(),
27 => twenty_seven::main(),
// 28 => twenty_eight::main(),
// 29 => twenty_nine::main(),
// 30 => thirty::main(),