Arithmetic operators - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of mixed arithmetic operations

What is the output of this C code snippet?

C

#include <stdio.h>

int main() {
    int a = 5, b = 2;
    int result = a / b * b + a % b;
    printf("%d\n", result);
    return 0;
}

A 6

B 7

C 5

D 4

💡 Hint

Remember integer division truncates the decimal part.

The expression evaluates as (5 / 2) * 2 + (5 % 2) = 2 * 2 + 1 = 5.

❓ Predict Output

intermediate

Result of increment and multiplication

What will be printed by this C program?

C

#include <stdio.h>

int main() {
    int x = 3;
    int y = x++ * 2 + ++x;
    printf("%d\n", y);
    return 0;
}

A 10

B 11

C 13

D 12

💡 Hint

Remember post-increment uses the value before increment, pre-increment increments before use.

x starts as 3. x++ evaluates to 3 (x becomes 4 after), so x++ * 2 = 6. ++x then increments x to 5 and evaluates to 5. Thus, y = 6 + 5 = 11.

❓ Predict Output

advanced

Division and modulus with negative numbers

What is the output of this C code?

C

#include <stdio.h>

int main() {
    int a = -7, b = 3;
    printf("%d %d\n", a / b, a % b);
    return 0;
}

A -3 -1

B -3 2

C -2 1

D -2 -1

💡 Hint

In C, integer division truncates toward zero and modulus sign matches the dividend.

In C, -7 / 3 = -2 (truncates toward zero), and -7 % 3 = -1 (sign matches dividend).

🔧 Debug

advanced

Identify the error in arithmetic expression

What error will this C code produce when compiled?

C

#include <stdio.h>

int main() {
    int a = 10, b = 3;
    int c = a / b *;
    printf("%d\n", c);
    return 0;
}

A SyntaxError: missing operand after '*'

B SyntaxError: stray '*' in program

C SyntaxError: expected expression before ';'

D Compilation error: invalid operands to binary *

💡 Hint

Look carefully at the arithmetic expression syntax.

The expression 'a / b *;' is incomplete because '*' operator lacks a right operand, causing a syntax error.

🚀 Application

expert

Calculate the number of digits in an integer

Which code snippet correctly calculates the number of digits in a positive integer n using arithmetic operators only?

A

int count = 0;
while (n > 0) {
    n = n / 10;
    count++;
}

B

int count = 1;
while (n >= 10) {
    n = n % 10;
    count++;
}

C

int count = 0;
while (n != 0) {
    n = n * 10;
    count++;
}

D

int count = 0;
while (n > 0) {
    n = n - 10;
    count++;
}

💡 Hint

Dividing by 10 removes the last digit of a number.

Option A divides n by 10 repeatedly, counting how many times until n is 0, which equals the number of digits.

Practice

1. Which arithmetic operator in C gives the remainder of a division?

easy

A. The multiplication operator (*)

B. The division operator (/)

C. The modulus operator (%)

D. The subtraction operator (-)

Solution

1. Step 1: Understand the modulus operator

   The modulus operator (%) returns the remainder after division of two integers.

2. Step 2: Compare with other operators

   Division (/) returns the quotient, multiplication (*) multiplies, subtraction (-) subtracts.

3. Final Answer:

   The modulus operator (%) -> Option C

4. Quick Check:

   Remainder = % operator [OK]

Hint: Remainder means modulus (%) operator [OK]

Common Mistakes:

• Confusing division (/) with modulus (%)
• Thinking multiplication (*) gives remainder
• Using subtraction (-) for remainder

2. Which of the following is the correct syntax to multiply two integers a and b in C?

easy

A. a x b;

B. a * b;

C. a . b;

D. a & b;

Solution

1. Step 1: Recall multiplication operator in C

   The multiplication operator in C is the asterisk symbol (*).

2. Step 2: Check each option

   a * b; uses '*', which is correct. Options B, C, D use invalid symbols for multiplication.

3. Final Answer:

   a * b; -> Option B

4. Quick Check:

   Multiply with * operator [OK]

Hint: Use * for multiplication in C [OK]

Common Mistakes:

• Using 'x' instead of '*' for multiplication
• Using '.' or '&' mistakenly
• Missing semicolon at end

3. What is the output of this C code?

int a = 7, b = 3;
printf("%d", a / b);

medium

A. 1

B. 2.333333

C. 3

D. 2

Solution

1. Step 1: Understand integer division in C

   Both a and b are integers, so a / b performs integer division, which discards the decimal part.

2. Step 2: Calculate 7 / 3 using integer division

   7 divided by 3 is 2.333..., but integer division returns 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Integer division truncates decimals [OK]

Hint: Integer division drops decimals, result is integer [OK]

Common Mistakes:

• Expecting floating-point result
• Confusing remainder with division result
• Using wrong format specifier in printf

4. Find the error in this C code snippet:

int x = 10;
int y = 3;
int z = x / y.0;
printf("%d", z);

medium

A. Using a floating-point number in integer division

B. Incorrect format specifier in printf

C. Missing semicolon after variable declaration

D. Division by zero error

Solution

1. Step 1: Analyze the division expression

   The expression x / y.0 uses y.0 which attempts a floating-point literal but causes invalid syntax, as x and y are integers.

2. Step 2: Understand the syntax error

   Using y.0 causes a compilation error due to invalid syntax in the integer division context.

3. Final Answer:

   Using a floating-point number in integer division -> Option A

4. Quick Check:

   Integer division must use integers only [OK]

Hint: Integer division needs integer operands only [OK]

Common Mistakes:

• Thinking printf format is wrong
• Ignoring the decimal in y.0
• Assuming division by zero

5. You want to calculate the average of two integers a and b in C and get a floating-point result. Which code snippet correctly does this?

hard

A. float avg = (a + b) / 2.0;

B. float avg = (a + b) / 2;

C. int avg = (a + b) / 2.0;

D. float avg = a / b + 2;

Solution

1. Step 1: Understand integer vs floating-point division

   Dividing by 2 (integer) causes integer division, losing decimals. Dividing by 2.0 (double) forces floating-point division.

2. Step 2: Check each option for correct average calculation

   float avg = (a + b) / 2; does integer division then converts to float, losing decimals. float avg = (a + b) / 2.0; divides by 2.0, giving float result. int avg = (a + b) / 2.0; stores float result in int, losing decimals. float avg = a / b + 2; is incorrect formula.

3. Final Answer:

   float avg = (a + b) / 2.0; -> Option A

4. Quick Check:

   Divide by 2.0 for float average [OK]

Hint: Divide by 2.0 to get float average [OK]

Common Mistakes:

• Dividing by integer 2 causing integer division
• Storing float result in int variable
• Wrong formula for average