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 '<' causes wrong logic.

Checking for num == 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 * i [1] 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 '<' fails for perfect square composites like 9.

Using '>' or '>=' reverses the loop condition.

The loop should run while i squared is less than or equal to num.

3fill in blank

hard

Fix the error in the if condition that checks divisibility.

DSA C++

if (num [1] 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 '/' checks division but does not check remainder.

Using '*' or '-' causes syntax or logic errors.

The modulus operator '%' checks if num is divisible by i with no remainder.

4fill in blank

hard

Fill both blanks to complete the function header and return type for checking prime.

DSA C++

[1] isPrime(int num) {
    [2];
}

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

Abool

Breturn true

Cint

Dreturn false

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'int' as return type instead of 'bool'.

Forgetting to return true at the end of the function.

The function returns a boolean value, so use 'bool' as return type and 'return true' if no divisor found.

5fill in blank

hard

Fill all three blanks to complete the prime check function with input and output.

DSA C++

#include <iostream>

int main() {
    int num;
    std::cin >> [1];
    if (isPrime([2])) {
        std::cout << [3] << std::endl;
    } else {
        std::cout << "Not prime" << std::endl;
    }
    return 0;
}

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

Anum

C"Prime"

Dnumber

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing errors.

Printing variable name instead of string "Prime".

Read input into 'num', check isPrime(num), and print "Prime" if true.

Practice

1. What does it mean if a number is 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 a prime number has 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 means divisible only by 1 and itself [OK]

Common Mistakes:

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

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

easy

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

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/2; i++) { if (n % i == 0) return true; } return false;

Solution

1. Step 1: Analyze loop range for prime check

   We check divisors from 2 up to but not including n to find any divisor.

2. Step 2: Check correctness of each option

   for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; correctly checks divisors from 2 to n-1 and returns false if divisible, 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 first divisor found, which is incorrect logic.

3. Final Answer:

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

4. Quick Check:

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

Hint: Check divisors from 2 to n-1 only [OK]

Common Mistakes:

• Checking divisibility by 1 or n itself
• Looping beyond n
• Returning true too early

3. What is the output of the following C++ code?

int n = 11;
bool isPrime = true;
for (int i = 2; i <= n / 2; i++) {
    if (n % i == 0) {
        isPrime = false;
        break;
    }
}
std::cout << (isPrime ? "Prime" : "Not Prime") << std::endl;

medium

A. Runtime Error

B. Not Prime

C. Compilation Error

D. Prime

Solution

1. Step 1: Understand the loop and condition

   The loop checks divisors from 2 to n/2 (which is 5 for n=11). It tests if 11 is divisible by any number in this range.

2. Step 2: Check divisibility for each i

   11 % 2 != 0, 11 % 3 != 0, 11 % 4 != 0, 11 % 5 != 0. No divisor found, so isPrime remains true.

3. Final Answer:

   Prime -> Option D

4. Quick Check:

   11 has no divisors except 1 and itself [OK]

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

Common Mistakes:

• Looping up to n instead of n/2
• Not breaking loop on divisor found
• Misinterpreting output

4. Find the error in this C++ code that checks if n is prime:

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

medium

A. Loop should start from 1, not 2.

B. Returns true when a divisor is found, should return false instead.

C. Function should return true for n <= 1.

D. No error, code is correct.

Solution

1. Step 1: Analyze return values on divisor found

   The code returns true immediately when a divisor is found, which means it claims the number is prime incorrectly.

2. Step 2: Correct logic for prime check

   When a divisor is found, the function should return false (not prime). If no divisor is found, return true.

3. Final Answer:

   Returns true when a divisor is found, should return false instead. -> Option B

4. Quick Check:

   Return false on divisor found [OK]

Hint: Return false when divisor found, true otherwise [OK]

Common Mistakes:

• Returning true on divisor found
• Loop starting at 1
• Not handling n <= 1 correctly

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

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

hard

A. Incorrect: returns true inside the loop when divisor found.

B. Incorrect: loop should go up to n/2, not sqrt(n).

C. Correct: checks divisors up to sqrt(n) and returns false if divisible.

D. Incorrect: should start loop from 1 instead of 2.

Solution

1. Step 1: Understand optimization using square root

   Any divisor larger than sqrt(n) pairs with a smaller divisor, so checking up to sqrt(n) is enough.

2. Step 2: Verify code logic

   The loop runs from 2 to sqrt(n), returns false if divisor found, else true after loop ends. This is correct and efficient.

3. Final Answer:

   Correct: checks divisors up to sqrt(n) and returns false if divisible. -> Option C

4. Quick Check:

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

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

Common Mistakes:

• Looping up to n/2 instead of sqrt(n)
• Starting loop at 1
• Returning true inside loop on divisor found