Check if Number is Prime in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a number is less than 2, which means it is not prime.

DSA C#

if (num [1] 2) {
    return false;
}

Drag options to blanks, or click blank then click option'

A==

B>

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' will give wrong results.

Checking for equality to 2 only misses numbers less than 2.

If the number is less than 2, it is not prime.

2fill in blank

medium

Complete the for loop to check divisors from 2 up to the square root of the number.

DSA C#

for (int i = 2; i [1] Math.Sqrt(num); i++) {
    if (num % i == 0) {
        return false;
    }
}

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '>=' will cause the loop to never run.

Using '<' will miss checking divisors equal to the square root (e.g., for perfect squares).

We need to check divisors up to the square root of the number (inclusive), so use '<='.

3fill in blank

hard

Fix the error in the return statement to correctly indicate the number is prime after the loop.

DSA C#

return [1];

Drag options to blanks, or click blank then click option'

Afalse

Btrue

Cnum

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning false always will mark all numbers as not prime.

Returning the number itself is incorrect.

After checking all divisors, if none divide the number, it is prime, so return true.

4fill in blank

hard

Fill both blanks to complete the method signature and the return type for checking prime.

DSA C#

public static [1] IsPrime(int num) {
    [2]

Drag options to blanks, or click blank then click option'

Abool

Bint

Creturn true;

Dreturn num;

Attempts:

3 left

💡 Hint

Common Mistakes

Using int as return type is incorrect for true/false results.

Returning the number itself does not indicate primality.

The method returns a boolean indicating if the number is prime, so the return type is bool and it returns true or false.

5fill in blank

hard

Fill all three blanks to complete the prime check method with initialization, loop condition, and return statement.

DSA C#

public static bool IsPrime(int num) {
    if (num [1] 2) return false;
    for (int i = 2; i [2] Math.Sqrt(num); i++) {
        if (num % i == 0) return false;
    }
    return [3];
}

Drag options to blanks, or click blank then click option'

A<

B<=

Ctrue

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' in the loop condition causes no iterations.

Returning false at the end marks all numbers as not prime.

The method returns false if number is less than 2, loops from 2 up to the square root, and returns true if no divisor found.

Practice

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

easy

A. It is an even number greater than 2.

B. It is divisible by 2 only.

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

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

   It has exactly two distinct positive divisors: 1 and itself. matches this definition exactly, while others do not.

3. Final Answer:

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

4. Quick Check:

   Prime number = two divisors [OK]

Hint: Prime means only divisible by 1 and itself [OK]

Common Mistakes:

• Thinking even numbers are always prime
• Confusing prime with odd numbers
• Believing prime numbers have multiple divisors

2. Which of the following C# 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 false; } return true;

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

C. for (int i = 2; i < n; i++) { if (n % i == 1) return false; } return true;

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

Solution

1. Step 1: Analyze the loop range and condition

   for (int i = 2; i <= n / 2; i++) { if (n % i == 0) return false; } return true; checks divisors from 2 up to n/2, which is sufficient to find factors for primality.

2. Step 2: Check the modulo condition and return values

   for (int i = 2; i <= n / 2; i++) { if (n % i == 0) return false; } return true; returns false if a divisor is found, otherwise true, correctly identifying primes.

3. Final Answer:

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

4. Quick Check:

   Loop 2 to n/2 and check divisibility [OK]

Hint: Check divisors from 2 to n/2 for efficiency [OK]

Common Mistakes:

• Starting loop from 1 instead of 2
• Checking modulo equals 1 instead of 0
• Returning true inside the loop incorrectly

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

bool IsPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i <= Math.Sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
Console.WriteLine(IsPrime(11));

medium

A. False

B. Runtime error

C. Compile error

D. True

Solution

1. Step 1: Understand the prime check logic

   The function returns false if n <= 1, else checks divisors from 2 to sqrt(n). For 11, sqrt(11) ~ 3.3, so loop checks i=2,3.

2. Step 2: Check divisibility for n=11

   11 % 2 != 0 and 11 % 3 != 0, so loop completes without returning false, function returns true.

3. Final Answer:

   True -> Option D

4. Quick Check:

   11 is prime, so output is True [OK]

Hint: Check divisors only up to square root [OK]

Common Mistakes:

• Not using Math.Sqrt and checking too many numbers
• Returning false for prime numbers
• Confusing output with string instead of boolean

4. Identify the error in this C# code snippet for checking if n is prime:

bool IsPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < n / 2; i++) {
    if (n % i == 0) return false;
  }
  return true;
}

medium

A. Loop should use <= instead of < to include n/2

B. Check for n <= 1 should be after the loop

C. Return true should be inside the loop

D. Loop should run up to Math.Sqrt(n) for efficiency

Solution

1. Step 1: Test with n=4

   n/2 = 2 (integer division), loop i=2 < 2 is false, loop skipped, returns true incorrectly (4 is not prime).

2. Step 2: Identify the fix

   Change to i <= n/2 includes i=2 when n/2=2, checking 4 % 2 == 0 and returning false.

3. Final Answer:

   Loop should use <= instead of < to include n/2 -> Option A

4. Quick Check:

   For n=4, i <= n/2 checks divisor 2 [OK]

Hint: Use <= n/2 to include boundary divisor check [OK]

Common Mistakes:

• Using < instead of <= for loop upper bound
• Returning true inside the loop prematurely
• Checking n <= 1 after the loop

5. You want to check if a large number n is prime efficiently. Which approach below is best to implement in C#?

hard

A. Check divisibility from 2 to n-1 using a for loop

B. Check divisibility from 2 to Math.Sqrt(n) using a for loop

C. Check divisibility from 2 to n/2 using a for loop

D. Check divisibility only by 2 and 3

Solution

1. Step 1: Understand the efficiency of divisor checking

   Checking all numbers up to n-1 or n/2 is slow for large n.

2. Step 2: Use mathematical property to limit checks

   Any factor larger than sqrt(n) pairs with a smaller factor, so checking up to sqrt(n) suffices.

3. Step 3: Evaluate options

   Check divisibility from 2 to Math.Sqrt(n) using a for loop uses sqrt(n) limit, which is efficient and correct. Check divisibility only by 2 and 3 is incomplete as primes can have other factors.

4. Final Answer:

   Check divisibility from 2 to Math.Sqrt(n) using a for loop -> Option B

5. Quick Check:

   Use sqrt(n) limit for best prime check [OK]

Hint: Limit divisor checks to square root of number [OK]

Common Mistakes:

• Checking too many numbers causing slow code
• Only checking 2 and 3 misses other factors
• Not handling edge cases like n <= 1