Welcome to collectivesolver - Programming & Software Q&A with code examples. A website with trusted programming answers. All programs are tested and work.

Contact: aviboots(AT)netvision.net.il

Semrush - keyword research tool

Create your online store today with Shopify

Turn ChatGPT, Claude, Gemini, And CoPilot Into Your Personal Assistant, Business Coach, Content Creator, And More

AFFILIATE MARKETING Your all-in-one performance engine Manage affiliates, creators, and customer referrals in one unified platform—turning every partnership into measurable growth

Secure & Reliable Web Hosting, Free Domain, Free SSL, 1-Click WordPress Install, Expert 24/7 Support

Disclosure: My content contains affiliate links.

43,181 questions

56,073 answers

573 users

How to find the largest prime factor of a number in Rust

2 Answers

0 votes
fn largest_prime_factor(mut n: i64) -> i64 {
    let mut div = 2;
    let mut max_p_fact = -1;

    while n != 0 {
        if n % div != 0 {
            div += 1;
        } else {
            max_p_fact = n;
            n /= div; // Integer division
            if n == 1 {
                break;
            }
        }
    }

    max_p_fact
}

fn main() {
    println!("{}", largest_prime_factor(124));  // 2 x 2 x 31
    println!("{}", largest_prime_factor(288));  // 2 x 2 x 2 x 2 x 2 x 3 x 3
    println!("{}", largest_prime_factor(1288)); // 2 x 2 x 2 x 7 x 23
    println!("{}", largest_prime_factor(100000000)); // 2, 2, 2, 2, 2, 2, 2, 2, 5, 5, 5, 5, 5, 5, 5, 5
}



/*
run:

31
3
23
5

*/

 



answered Apr 16, 2025 by avibootz
0 votes
fn largest_prime_factor(mut n: i64) -> i64 {
    let mut largest: i64 = -1;

    // Step 1: Remove all factors of 2
    while n % 2 == 0 {
        largest = 2;
        n /= 2;
    }

    // Step 2: Try odd factors from 3 to sqrt(n)
    let mut i: i64 = 3;
    while i * i <= n {
        while n % i == 0 {
            largest = i;
            n /= i;
        }
        i += 2;
    }

    // Step 3: If n is still > 2, it is a prime number
    if n > 2 {
        largest = n;
    }

    largest
}

fn main() {
    let n1: i64 = 124; // 2 x 2 x 31
    let n2: i64 = 288; // 2 x 2 x 2 x 2 x 2 x 3 x 3
    let n3: i64 = 1288; // 2 x 2 x 2 x 7 x 23
    let n4: i64 = 100000000; // 2, 2, 2, 2, 2, 2, 2, 2, 5, 5, 5, 5, 5, 5, 5, 5
    let n5: i64 = 600851475143; // 71, 893, 1471, 6857

    println!("Largest prime factor: {}", largest_prime_factor(n1));
    println!("Largest prime factor: {}", largest_prime_factor(n2));
    println!("Largest prime factor: {}", largest_prime_factor(n3));
    println!("Largest prime factor: {}", largest_prime_factor(n4));
    println!("Largest prime factor: {}", largest_prime_factor(n5));
}


/*
run:

Largest prime factor: 31
Largest prime factor: 3
Largest prime factor: 23
Largest prime factor: 5
Largest prime factor: 6857

*/

 



answered Apr 23 by avibootz
...