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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of prime check for number 29

What is the output of the following Java code that checks if 29 is prime?

DSA Java

public class PrimeCheck {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println(isPrime(29));
    }
}

A Runtime exception

B Compilation error

C true

D false

💡 Hint

Check if 29 has any divisors other than 1 and itself.

The code checks divisors from 2 up to the square root of 29. Since 29 is prime, no divisor is found, so it returns true.

❓ Predict Output

intermediate

Output of prime check for number 1

What is the output of the following Java code that checks if 1 is prime?

DSA Java

public class PrimeCheck {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println(isPrime(1));
    }
}

A false

B true

C Compilation error

D Runtime exception

💡 Hint

Remember prime numbers are greater than 1.

The code returns false immediately for numbers less than or equal to 1, so 1 is not prime.

🔧 Debug

advanced

Identify the error in prime check code

What error will this Java code produce when checking if 10 is prime?

DSA Java

public class PrimeCheck {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i <= n / 2; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println(isPrime(10));
    }
}

A Compilation error

B false

C true

D Runtime exception

💡 Hint

Check the loop condition and divisor range.

The code checks divisors from 2 up to n/2 (inclusive). For 10, it finds divisor 2 and returns false correctly.

❓ Predict Output

advanced

Output of prime check for number 2

What is the output of the following Java code that checks if 2 is prime?

DSA Java

public class PrimeCheck {
    public static boolean isPrime(int n) {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) return false;
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println(isPrime(2));
    }
}

A Runtime exception

B false

C Compilation error

D true

💡 Hint

2 is the smallest prime number.

The loop condition i * i <= n is false for i=2 (4 <= 2 is false), so loop does not run and returns true.

🧠 Conceptual

expert

Why check divisors only up to square root of n?

Why is it enough to check divisors only up to the square root of a number n when checking if n is prime?

A Because any divisor larger than the square root would have a corresponding divisor smaller than the square root.

B Because checking beyond the square root causes runtime errors.

C Because numbers larger than the square root cannot divide n evenly.

D Because prime numbers only have divisors less than their square root.

💡 Hint

Think about pairs of factors.

If n = a * b, then one of a or b must be less than or equal to sqrt(n). So checking beyond sqrt(n) is redundant.

Practice

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

easy

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

B. It is divisible by 2 only.

C. It is an even number greater than 2.

D. It has more than two divisors.

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 that prime numbers have exactly two distinct positive divisors: 1 and itself.

3. Final Answer:

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

4. Quick Check:

   Prime number = 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 Java code snippets correctly checks if a number n is prime?

easy

A. for (int i = 2; i < n / 2; i++) { if (n % i == 0) return true; } return false;

B. for (int i = 2; i <= n; i++) { if (n % i == 0) return false; } return true;

C. for (int i = 1; i < n; i++) { if (n % i == 0) return false; } return true;

D. for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true;

Solution

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

   We check divisors from 2 up to but not including n to see if n is divisible by any number other than 1 and itself.

2. Step 2: Check each option's logic

   for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; correctly loops from 2 to n-1 and returns false if a divisor is found, else true. for (int i = 2; i <= n; i++) { if (n % i == 0) return false; } return true; incorrectly includes n itself, causing false negatives. for (int i = 1; i < n; i++) { if (n % i == 0) return false; } return true; starts from 1, which divides all numbers. for (int i = 2; i < n / 2; i++) { if (n % i == 0) return true; } return false; returns true on finding a divisor but should return false.

3. Final Answer:

   for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; -> Option D

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:

• Checking divisibility including n itself
• Starting loop from 1 instead of 2
• Returning true when divisor found

3. What is the output of the following Java code when n = 11?

boolean isPrime = true;
for (int i = 2; i <= Math.sqrt(n); i++) {
    if (n % i == 0) {
        isPrime = false;
        break;
    }
}
System.out.println(isPrime);

medium

A. true

B. false

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Understand the loop range and condition

   The loop runs from 2 to the square root of 11 (approximately 3.3), so i takes values 2 and 3.

2. Step 2: Check divisibility for i=2 and i=3

   11 % 2 != 0 and 11 % 3 != 0, so no divisor found, isPrime remains true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   11 is prime, so output is true [OK]

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

Common Mistakes:

• Looping up to n instead of sqrt(n)
• Forgetting to break loop on divisor found
• Misunderstanding boolean flag usage

4. Find the error in the following Java code that checks if n is prime:

boolean isPrime = true;
for (int i = 2; i <= n / 2; i++) {
    if (n % i == 0) {
        isPrime = true;
        break;
    }
}
System.out.println(isPrime);

medium

A. The loop should run until i < n, not n / 2.

B. The flag isPrime should be set to false when divisor found, not true.

C. The loop variable i should start from 1, not 2.

D. There is no error; the code works correctly.

Solution

1. Step 1: Analyze the flag update inside the loop

   When a divisor is found, isPrime should be set to false, but here it is incorrectly set to true.

2. Step 2: Check loop range and starting point

   The loop from 2 to n/2 is acceptable for checking divisors, so no error there.

3. Final Answer:

   The flag isPrime should be set to false when divisor found, not true. -> Option B

4. Quick Check:

   Set isPrime = false on divisor found [OK]

Hint: Set prime flag false when divisor found [OK]

Common Mistakes:

• Setting isPrime true inside divisor check
• Looping from 1 instead of 2
• Ignoring break after finding divisor

5. You want to write a Java method isPrime(int n) that returns true if n is prime and false otherwise. Which of the following improvements makes the method more efficient for large n?

hard

A. Check divisibility by all numbers from 2 to n-1 without any break.

B. Check divisibility only by even numbers.

C. Check divisibility only up to the square root of n instead of n-1.

D. Check divisibility by numbers greater than n/2 only.

Solution

1. Step 1: Understand the divisor checking range

   Any divisor larger than the square root of n would have a corresponding divisor smaller than the square root, so checking beyond sqrt(n) is unnecessary.

2. Step 2: Evaluate options for efficiency

   Check divisibility only up to the square root of n instead of n-1. reduces the number of checks drastically, improving efficiency. Check divisibility by all numbers from 2 to n-1 without any break. is inefficient. Check divisibility only by even numbers. is wrong because even numbers except 2 are not prime. Check divisibility by numbers greater than n/2 only. misses smaller divisors.

3. Final Answer:

   Check divisibility only up to the square root of n instead of n-1. -> Option C

4. Quick Check:

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

Hint: Only check divisors up to sqrt(n) for speed [OK]

Common Mistakes:

• Checking all numbers up to n-1
• Checking only even numbers
• Ignoring divisor pairs below sqrt(n)