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.