Check if Number is Even or Odd Using Bits in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code snippet?

Given the code below, what will be printed when the input number is 10?

DSA C

#include <stdio.h>

int main() {
    int num = 10;
    if (num & 1) {
        printf("Odd\n");
    } else {
        printf("Even\n");
    }
    return 0;
}

A Even

B Odd

C 10

D Compilation error

💡 Hint

Check the least significant bit of the number using bitwise AND with 1.

The bitwise AND operation num & 1 checks the least significant bit. For 10 (binary 1010), the last bit is 0, so it is even.

❓ Predict Output

intermediate

What is the output when num = 7?

Analyze the code below and determine the output when num is 7.

DSA C

#include <stdio.h>

int main() {
    int num = 7;
    if ((num & 1) == 1) {
        printf("Odd\n");
    } else {
        printf("Even\n");
    }
    return 0;
}

A Runtime error

B Odd

C 7

D Even

💡 Hint

The expression (num & 1) checks if the last bit is set.

7 in binary is 111, so the last bit is 1, making it odd.

🧠 Conceptual

advanced

Why does using bitwise AND with 1 check if a number is even or odd?

Select the best explanation for why the expression (num & 1) can determine if num is even or odd.

A Because bitwise AND with 1 shifts the number right by one bit.

B Because it converts the number to its negative equivalent.

C Because the least significant bit of an odd number is always 1, and for even numbers it is 0.

D Because it counts the total number of 1 bits in the number.

💡 Hint

Think about the binary representation of even and odd numbers.

In binary, odd numbers always end with 1, so ANDing with 1 isolates that bit to check parity.

❓ Predict Output

advanced

What is the output of this code snippet?

What will be printed when the input number is -3?

DSA C

#include <stdio.h>

int main() {
    int num = -3;
    if (num & 1) {
        printf("Odd\n");
    } else {
        printf("Even\n");
    }
    return 0;
}

A Odd

B Even

C 0

D Undefined behavior

💡 Hint

Consider how negative numbers are represented in two's complement.

In two's complement, negative odd numbers still have the least significant bit set to 1, so the output is Odd.

🔧 Debug

expert

What error does this code produce?

Identify the error in the following code snippet.

DSA C

#include <stdio.h>

int main() {
    int num = 4;
    if (num & 1 == 1) {
        printf("Odd\n");
    } else {
        printf("Even\n");
    }
    return 0;
}

A Runtime error due to invalid operation

B Syntax error due to missing parentheses

C Always prints Even regardless of num

D Always prints Odd regardless of num

💡 Hint

Check operator precedence between & and ==.

The expression num & 1 == 1 is parsed as num & (1 == 1), which is num & 1, so the condition is true for any num with last bit 1. For num=4 (binary 100), num & 1 is 0, so condition false, prints Even. But for num=5, it prints Odd. The problem is subtle: the condition is correct but can be confusing without parentheses.