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.