Modular Arithmetic Basics in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the remainder of 10 divided by 3.

DSA C

int result = 10 [1] 3;
printf("%d\n", result);

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

A+

B*

C/

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using / instead of % will give the quotient, not remainder.

Using * or + will not compute remainder.

The % operator gives the remainder of division in C.

2fill in blank

medium

Complete the code to compute (7 + 5) modulo 6.

DSA C

int sum = (7 + 5) [1] 6;
printf("%d\n", sum);

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

A%

B*

C/

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using / will give quotient, not remainder.

Using * or - will not compute modulo.

The modulo operator % gives the remainder after division, so (7+5)%6 = 12%6 = 0.

3fill in blank

hard

Fix the error in the code to correctly compute (a * b) modulo m.

DSA C

int mod_mul(int a, int b, int m) {
    return (a [1] b) % m;
}

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

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - will add or subtract instead of multiply.

Using / will cause division error or wrong result.

To multiply a and b before modulo, use the * operator.

4fill in blank

hard

Fill both blanks to compute (x - y) modulo n correctly, handling negative results.

DSA C

int mod_sub(int x, int y, int n) {
    int diff = (x [1] y) % n;
    return diff [2] 0 ? diff + n : diff;
}

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

A-

B<

C>

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - for subtraction.

Checking > 0 instead of < 0 for negative.

Subtract y from x, then if result is negative (less than 0), add n to keep it positive modulo.

5fill in blank

hard

Fill all three blanks to create a function that computes (a^b) modulo m using fast exponentiation.

DSA C

int mod_pow(int a, int b, int m) {
    int result = 1;
    a = a [1] m;
    while (b > 0) {
        if (b [2] 1) result = (result * a) % m;
        a = (a * a) % m;
        b = b [3] 1;
    }
    return result;
}

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

A%

B&

C>>

D<<

Attempts:

3 left

💡 Hint

Common Mistakes

Using / instead of >> for shifting.

Using + or - instead of & for bit check.

Not applying modulo after multiplication.

Use modulo to keep a in range, bitwise & 1 to check odd b, and right shift >> 1 to divide b by 2.

Practice

1. What is the result of 7 % 3 in modular arithmetic?

easy

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Understand the modulo operation

   The modulo operator (%) gives the remainder when the first number is divided by the second.

2. Step 2: Calculate 7 % 3

   7 divided by 3 is 2 with a remainder of 1, but since 7 % 3 means remainder, it is 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   7 % 3 = 1 [OK]

Hint: Modulo gives remainder after division [OK]

Common Mistakes:

• Confusing quotient with remainder
• Using division instead of modulo
• Mixing up dividend and divisor

2. Which of the following is the correct syntax to compute the remainder of a divided by b in C?

easy

A. a mod b

B. a % b

C. mod(a, b)

D. a / b

Solution

1. Step 1: Recall C syntax for modulo

   In C, the modulo operator is represented by the percent sign '%'.

2. Step 2: Identify correct syntax

   The correct way to get remainder is a % b. Other options are invalid syntax or division.

3. Final Answer:

   a % b -> Option B

4. Quick Check:

   Modulo operator in C is % [OK]

Hint: Use % operator for modulo in C [OK]

Common Mistakes:

• Using 'mod' keyword
• Using division operator '/' instead
• Calling a function mod() which doesn't exist in C

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

#include 
int main() {
    int x = 14;
    int y = 5;
    printf("%d", (x + y) % y);
    return 0;
}

medium

A. 4

B. 0

C. 1

D. 5

Solution

1. Step 1: Calculate (x + y)

   x = 14, y = 5, so x + y = 19.

2. Step 2: Calculate (x + y) % y

   19 % 5 is the remainder when 19 is divided by 5. 5 * 3 = 15, remainder 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   (14 + 5) % 5 = 4 [OK]

Hint: Add first, then apply modulo [OK]

Common Mistakes:

• Applying modulo before addition
• Confusing remainder with quotient
• Incorrect operator precedence

4. Identify the error in the following C code snippet that attempts to compute a % b:

int a = 10;
int b = 0;
int c = a % b;
printf("%d", c);

medium

A. Syntax error in modulo operator

B. Variable c is not declared

C. Division by zero error

D. No error, code runs fine

Solution

1. Step 1: Check divisor value

   Variable b is zero, which is the divisor in modulo operation.

2. Step 2: Understand modulo with zero divisor

   Modulo by zero is undefined and causes a runtime error (division by zero).

3. Final Answer:

   Division by zero error -> Option C

4. Quick Check:

   Modulo divisor cannot be zero [OK]

Hint: Never modulo by zero; it causes runtime error [OK]

Common Mistakes:

• Ignoring zero divisor
• Assuming modulo by zero returns zero
• Confusing syntax error with runtime error

5. Given the expression (a * b) % m, which of the following is a correct way to compute it without causing integer overflow in C when a and b are large integers?

hard

A. Use (a % b) % m

B. Use (a * b) % m directly

C. Use (a + b) % m

D. Use ((a % m) * (b % m)) % m

Solution

1. Step 1: Understand overflow risk

   Multiplying large integers a * b can overflow before modulo is applied.

2. Step 2: Apply modular multiplication property

   Using ((a % m) * (b % m)) % m keeps intermediate values smaller, preventing overflow.

3. Final Answer:

   ((a % m) * (b % m)) % m -> Option D

4. Quick Check:

   Modular multiplication reduces overflow risk [OK]

Hint: Modulo each operand before multiply to avoid overflow [OK]

Common Mistakes:

• Multiplying before modulo causing overflow
• Using addition instead of multiplication
• Incorrect modulo order