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 C# code using bitwise AND to check even or odd?

Look at the code below. What will it print when number = 7?

DSA C#

int number = 7;
string result = (number & 1) == 0 ? "Even" : "Odd";
Console.WriteLine(result);

A "Even"

B 1

C "Odd"

D 0

💡 Hint

The bitwise AND with 1 checks the last bit of the number.

The expression (number & 1) checks if the last bit is 1 (odd) or 0 (even). Since 7 in binary ends with 1, it is odd.

❓ Predict Output

intermediate

What is the output when number = 12 in this bitwise even/odd check?

Check the output of this code snippet:

DSA C#

int number = 12;
string result = (number & 1) == 0 ? "Even" : "Odd";
Console.WriteLine(result);

A "Odd"

B 0

C 12

D "Even"

💡 Hint

Even numbers have last bit 0 in binary.

12 in binary ends with 0, so (12 & 1) == 0 is true, printing "Even".

🔧 Debug

advanced

What error occurs in this C# code checking even/odd with bits?

Identify the error when running this code:

DSA C#

int number = 5;
string result = (number & 1) == 0 ? "Even" : "Odd";
Console.WriteLine(result);

A SyntaxError: missing semicolon

B TypeError: cannot apply & to int

C No error, prints "Odd"

D RuntimeException: null reference

💡 Hint

Check punctuation carefully in C# statements.

The first line is missing a semicolon at the end, causing a syntax error.

❓ Predict Output

advanced

What is the output of this code that uses bitwise XOR to check even or odd?

What does this print when number = 10?

DSA C#

int number = 10;
string result = ((number ^ 1) == number + 1) ? "Odd" : "Even";
Console.WriteLine(result);

A "Even"

B "Odd"

C 10

D 11

💡 Hint

XOR with 1 flips the last bit.

For even 10, 10 ^ 1 = 11 == 10 + 1, so condition true, prints "Odd".

❓ Predict Output

expert

What is the output of this C# code that uses bitwise operations to count even numbers in an array?

Given the array, how many even numbers does this code count?

DSA C#

int[] numbers = {3, 4, 7, 8, 10};
int count = 0;
foreach (int num in numbers) {
  if ((num & 1) == 0) {
    count++;
  }
}
Console.WriteLine(count);

A 3

B 2

C 4

D 5

💡 Hint

Count numbers whose last bit is 0.

4, 8, and 10 are even, so count is 3.

Practice

1. Which bit operation can be used to check if a number is even or odd in C#?

easy

A. Using bitwise OR with 1 (number | 1)

B. Using bitwise XOR with 1 (number ^ 1)

C. Using bitwise AND with 1 (number & 1)

D. Using bitwise NOT (~number)

Solution

1. Step 1: Understand bitwise AND with 1

   Bitwise AND with 1 checks the least significant bit of the number.

2. Step 2: Check least significant bit for even or odd

   If the least significant bit is 0, the number is even; if 1, it is odd.

3. Final Answer:

   Using bitwise AND with 1 (number & 1) -> Option C

4. Quick Check:

   Least significant bit check = number & 1 [OK]

Hint: Check last bit with & 1 to find even or odd fast [OK]

Common Mistakes:

• Using bitwise OR instead of AND
• Using XOR which toggles bits
• Using NOT which flips all bits

2. Which of the following C# code snippets correctly checks if num is odd using bits?

easy

A. if ((num & 1) == 0) { Console.WriteLine("Odd"); }

B. if ((num ^ 1) == 1) { Console.WriteLine("Odd"); }

C. if ((num | 1) == 1) { Console.WriteLine("Odd"); }

D. if ((num & 1) == 1) { Console.WriteLine("Odd"); }

Solution

1. Step 1: Understand bitwise AND check for odd

   Odd numbers have least significant bit 1, so (num & 1) == 1 means odd.

2. Step 2: Verify code logic

   if ((num & 1) == 1) { Console.WriteLine("Odd"); } prints "Odd" when (num & 1) equals 1, correctly identifying odd numbers.

3. Final Answer:

   if ((num & 1) == 1) { Console.WriteLine("Odd"); } -> Option D

4. Quick Check:

   Odd check = (num & 1) == 1 [OK]

Hint: Use (num & 1) == 1 to check odd numbers [OK]

Common Mistakes:

• Checking (num & 1) == 0 for odd (actually even)
• Using OR or XOR instead of AND
• Confusing equality check with assignment

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

int num = 14;
if ((num & 1) == 0)
    Console.WriteLine("Even");
else
    Console.WriteLine("Odd");

medium

A. Odd

B. Even

C. 14

D. Error

Solution

1. Step 1: Calculate bitwise AND of 14 and 1

   14 in binary is 1110; 1110 & 0001 = 0, so (num & 1) == 0 is true.

2. Step 2: Determine output based on condition

   Since condition is true, it prints "Even".

3. Final Answer:

   Even -> Option B

4. Quick Check:

   14 & 1 = 0 means Even [OK]

Hint: Even numbers have last bit 0, so (num & 1) == 0 [OK]

Common Mistakes:

• Assuming 14 is odd
• Confusing bitwise AND with OR
• Expecting numeric output instead of string

4. Find the error in this C# code that checks if a number is odd:

int num = 7;
if (num & 1 == 1)
    Console.WriteLine("Odd");
else
    Console.WriteLine("Even");

medium

A. Missing parentheses around (num & 1)

B. Using bitwise OR instead of AND

C. Incorrect variable type for num

D. No error, code is correct

Solution

1. Step 1: Analyze operator precedence

   In C#, equality (==) has higher precedence than bitwise AND (&), so 'num & 1 == 1' parses as 'num & (1 == 1)', which is 'num & 1'. This does not compare (num & 1) to 1 as intended.

2. Step 2: Identify the error

   Missing parentheses around (num & 1) causes incorrect evaluation. Correct code should be 'if ((num & 1) == 1)'.

3. Final Answer:

   Missing parentheses around (num & 1) -> Option A

4. Quick Check:

   Use parentheses to ensure correct precedence [OK]

Hint: C# precedence: == > & , so use parentheses to ensure correct evaluation [OK]

Common Mistakes:

• Misjudging precedence (thinking & > ==)
• Using bitwise OR instead of AND
• Assuming parentheses are not required

5. You want to write a C# method that returns "Even" or "Odd" for any integer using bitwise operations. Which of these implementations is correct and efficient?

public string CheckEvenOdd(int num) {
    // Options:
    // A) return (num & 1) == 0 ? "Even" : "Odd";
    // B) return (num | 1) == 0 ? "Even" : "Odd";
    // C) return (num ^ 1) == 0 ? "Even" : "Odd";
    // D) return (~num & 1) == 0 ? "Even" : "Odd";
}

hard

A. Option A

B. Option B

C. Option C

D. Option D

Solution

1. Step 1: Evaluate each option's logic

   Option A correctly checks if least significant bit is 0 for even; others use incorrect bitwise logic.

2. Step 2: Confirm correctness and efficiency

   Option A uses simple bitwise AND and ternary operator, efficient and clear.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Use (num & 1) == 0 for even check [OK]

Hint: Use (num & 1) == 0 ? "Even" : "Odd" for quick check [OK]

Common Mistakes:

• Using OR or XOR which give wrong results
• Using NOT which flips bits incorrectly
• Ignoring ternary operator for return