Check if Number is Prime in DSA Rust - Build from Scratch

Check if Number is Prime

📖 Scenario: Imagine you are a security guard checking if a number is special. A prime number is special because it can only be divided evenly by 1 and itself. You will write a program to check if a number is prime.

🎯 Goal: You will build a simple Rust program that checks if a given number is prime and prints the result.

📋 What You'll Learn

Create a variable called number with the value 29

Create a variable called is_prime and set it to true

Use a for loop with variable i from 2 to number - 1 to check divisibility

If number is divisible by i, set is_prime to false and break the loop

Print "29 is prime" if is_prime is true, otherwise print "29 is not prime"

💡 Why This Matters

🌍 Real World

Prime numbers are used in computer security, like encrypting messages to keep them safe.

💼 Career

Understanding prime numbers and loops is important for programming jobs that involve math, security, or algorithms.

1

Create the number variable

Create a variable called number and set it to 29

DSA Rust

# Your code here

Hint

Use let number = 29; inside the main function.

2

Create the is_prime variable

Create a variable called is_prime and set it to true inside the main function after number

DSA Rust

fn main() {
    let number = 29;
    # Your code here
}

Hint

Use let mut is_prime = true; to allow changing the value later.

3

Check divisibility with a for loop

Use a for loop with variable i from 2 to number - 1 inside the main function. Inside the loop, if number % i == 0, set is_prime = false and break the loop.

DSA Rust

fn main() {
    let number = 29;
    let mut is_prime = true;
    # Your code here
}

Hint

Use for i in 2..number and check divisibility with if number % i == 0.

4

Print the result

Print "29 is prime" if is_prime is true, otherwise print "29 is not prime" inside the main function after the loop.

DSA Rust

fn main() {
    let number = 29;
    let mut is_prime = true;
    for i in 2..number {
        if number % i == 0 {
            is_prime = false;
            break;
        }
    }
    # Your code here
}

Hint

Use if is_prime { println!("29 is prime"); } else { println!("29 is not prime"); }.

Practice

1. What does it mean for a number to be prime?

easy

A. It has more than two divisors.

B. It is divisible by 2 only.

C. It is an even number greater than 2.

D. It has exactly two distinct positive divisors: 1 and itself.

Solution

1. Step 1: Understand the definition of prime number

   A prime number is a number greater than 1 that has no divisors other than 1 and itself.

2. Step 2: Compare options with definition

   Only It has exactly two distinct positive divisors: 1 and itself. correctly states this definition.

3. Final Answer:

   It has exactly two distinct positive divisors: 1 and itself. -> Option D

4. Quick Check:

   Prime = two divisors [OK]

Hint: Prime numbers have only 1 and itself as divisors [OK]

Common Mistakes:

• Thinking even numbers are always prime
• Confusing prime with odd numbers
• Assuming 1 is prime

2. Which of the following Rust code snippets correctly checks if a number n is prime?

easy

A. for i in 1..n { if n % i == 0 { return false; } } true

B. for i in 2..n { if n % i == 0 { return false; } } true

C. for i in 2..=n { if n % i == 0 { return false; } } true

D. for i in 2..n { if n % i != 0 { return false; } } true

Solution

1. Step 1: Analyze the loop range for divisor checking

   We check divisors from 2 up to n-1 (exclusive of n) to find if any divides n.

2. Step 2: Check each option's loop and condition

   for i in 2..n { if n % i == 0 { return false; } } true correctly loops from 2 to n-1 and returns false if divisor found; others have errors like including n or wrong condition.

3. Final Answer:

   for i in 2..n { if n % i == 0 { return false; } } true -> Option B

4. Quick Check:

   Loop 2 to n-1 and check divisibility [OK]

Hint: Loop from 2 to n-1 to check divisors [OK]

Common Mistakes:

• Including n in the loop range
• Starting loop from 1 instead of 2
• Using wrong condition for divisibility

3. What is the output of this Rust function call: is_prime(11)?

fn is_prime(n: u32) -> bool {
    if n <= 1 { return false; }
    for i in 2..n {
        if n % i == 0 {
            return false;
        }
    }
    true
}

medium

A. true

B. false

C. panic at runtime

D. compile error

Solution

1. Step 1: Check input and loop behavior for n=11

   11 is greater than 1, so the function proceeds to loop from 2 to 10 checking divisibility.

2. Step 2: Test divisibility of 11 by numbers 2 to 10

   11 is not divisible by any number between 2 and 10, so loop completes without returning false.

3. Final Answer:

   true -> Option A

4. Quick Check:

   11 is prime, so function returns true [OK]

Hint: Prime numbers return true after no divisors found [OK]

Common Mistakes:

• Assuming 11 is not prime
• Confusing loop range
• Expecting runtime errors incorrectly

4. Identify the bug in this Rust function that checks if a number is prime:

fn is_prime(n: u32) -> bool {
    if n < 2 { return false; }
    for i in 2..=n {
        if n % i == 0 {
            return false;
        }
    }
    true
}

medium

A. Missing return type in function signature

B. Condition should be n <= 1 instead of n < 2

C. Loop includes n, causing always false for n > 1

D. Loop should start from 1 instead of 2

Solution

1. Step 1: Analyze the loop range 2..=n

   The loop includes n itself, so n % n == 0 is always true, causing immediate false return.

2. Step 2: Understand correct loop range

   Loop should be 2..n (exclusive of n) to avoid checking n against itself.

3. Final Answer:

   Loop includes n, causing always false for n > 1 -> Option C

4. Quick Check:

   Loop must exclude n to avoid false negatives [OK]

Hint: Exclude n from divisor loop to avoid false negatives [OK]

Common Mistakes:

• Including n in divisor checks
• Wrong loop start index
• Misunderstanding return conditions

5. You want to optimize prime checking for large numbers by only checking divisors up to the square root of n. Which Rust code snippet correctly implements this?

hard

A. for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true

B. for i in 2..n { if n % i == 0 { return false; } } true

C. for i in 2..=(n / 2) { if n % i == 0 { return false; } } true

D. for i in 1..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true

Solution

1. Step 1: Understand optimization using square root

   Checking divisors up to sqrt(n) is enough because larger factors pair with smaller ones.

2. Step 2: Analyze each option's loop range and correctness

   for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true correctly casts n to float, computes sqrt, casts back to u32, and loops up to that value.

3. Final Answer:

   for i in 2..=((n as f64).sqrt() as u32) { if n % i == 0 { return false; } } true -> Option A

4. Quick Check:

   Loop up to sqrt(n) for efficient prime check [OK]

Hint: Check divisors only up to sqrt(n) for efficiency [OK]

Common Mistakes:

• Checking divisors beyond sqrt(n)
• Starting loop from 1 instead of 2
• Incorrect casting causing errors