Check if Number is Prime in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of prime check function for 29

What is the output of this Rust code that checks if 29 is prime?

DSA Rust

fn is_prime(n: u32) -> bool {
    if n <= 1 {
        return false;
    }
    for i in 2..=((n as f64).sqrt() as u32) {
        if n % i == 0 {
            return false;
        }
    }
    true
}

fn main() {
    println!("{}", is_prime(29));
}

A false

B Compilation error

C true

D Runtime panic

💡 Hint

Recall that 29 is a prime number and the function returns true for primes.

The function correctly checks divisors up to the square root of 29. Since 29 has no divisors other than 1 and itself, it returns true.

❓ Predict Output

intermediate

Output of prime check function for 1

What is the output of this Rust code that checks if 1 is prime?

DSA Rust

fn is_prime(n: u32) -> bool {
    if n <= 1 {
        return false;
    }
    for i in 2..=((n as f64).sqrt() as u32) {
        if n % i == 0 {
            return false;
        }
    }
    true
}

fn main() {
    println!("{}", is_prime(1));
}

A Runtime panic

B false

C Compilation error

D true

💡 Hint

Remember that 1 is not considered a prime number.

The function returns false immediately for numbers less than or equal to 1, so it returns false for 1.

🔧 Debug

advanced

Identify the error in prime check function

What error does this Rust code produce when checking if 25 is prime?

DSA Rust

fn is_prime(n: u32) -> bool {
    if n <= 1 {
        return false;
    }
    for i in 2..(n as f64).sqrt() as u32 {
        if n % i == 0 {
            return false;
        }
    }
    true
}

fn main() {
    println!("{}", is_prime(25));
}

A true

B Runtime panic

C Compilation error

D false

💡 Hint

Check the range used in the for loop and whether it includes the square root value.

The range excludes the upper bound because it uses .. instead of ..=, so it misses checking divisor 5. But since 25 % 2 == 0 is false and 25 % 3 == 0 is false and 25 % 4 == 0 is false, the loop ends without finding divisor 5. So it incorrectly returns true. But the code actually returns true incorrectly.

❓ Predict Output

advanced

Output of prime check function for 0

What is the output of this Rust code that checks if 0 is prime?

DSA Rust

fn is_prime(n: u32) -> bool {
    if n <= 1 {
        return false;
    }
    for i in 2..=((n as f64).sqrt() as u32) {
        if n % i == 0 {
            return false;
        }
    }
    true
}

fn main() {
    println!("{}", is_prime(0));
}

A true

B Runtime panic

C Compilation error

D false

💡 Hint

Recall that 0 is not a prime number.

The function returns false immediately for numbers less than or equal to 1, so it returns false for 0.

🧠 Conceptual

expert

Why check divisors only up to square root in prime check?

Why does the prime checking function only test divisors up to the square root of the number?

A Because any divisor larger than the square root would have a matching smaller divisor already checked

B Because numbers larger than the square root are always prime

C Because checking beyond the square root causes runtime errors

D Because the square root is the largest possible prime factor

💡 Hint

Think about pairs of factors and how multiplication works.

If a number n has a divisor larger than its square root, it must be paired with a divisor smaller than the square root. So checking up to the square root is enough to find all divisors.

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