Modular Arithmetic Basics in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to compute the remainder of division.

DSA Java

int remainder = 17 [1] 5;

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

A%

B+

C/

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using / instead of %

Using * or + by mistake

The % operator gives the remainder when 17 is divided by 5, which is 2.

2fill in blank

medium

Complete the code to add two numbers and get the result modulo 10.

DSA Java

int result = (7 + 8) [1] 10;

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

A*

B-

C%

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using / instead of %

Using * or - by mistake

The % operator returns the remainder after division by 10, so (7 + 8) % 10 = 15 % 10 = 5.

3fill in blank

hard

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

DSA Java

int modProduct = (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 + instead of *

Using / or - by mistake

To compute the product modulo m, multiply a and b using * then take modulo m.

4fill in blank

hard

Fill both blanks to compute (x - y) modulo n correctly.

DSA Java

int modDiff = (x [1] y + n) [2] n;

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

A-

B+

C%

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to add n before modulo

Using * instead of %

Subtract y from x, add n to avoid negative results, then take modulo n.

5fill in blank

hard

Fill both blanks to create a map of numbers to their modulo 4 values for 1 to 5.

DSA Java

Map<Integer, Integer> modMap = new HashMap<>();
for (int i = 1; i <= 5; i++) {
    modMap.put(i, i [1] [2]);
}

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

B%

C4

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Adding operators to key

Using * instead of %

Wrong modulo number

Use i as key (no change), compute i % 4 as value.

Practice

1. What is the result of 17 % 5 in modular arithmetic?

easy

A. 0

B. 3

C. 5

D. 2

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 17 % 5

   Dividing 17 by 5 gives 3 with a remainder of 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   17 divided by 5 leaves remainder 2 [OK]

Hint: Modulo gives remainder after division [OK]

Common Mistakes:

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

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

easy

A. int result = a % b;

B. int result = a mod b;

C. int result = a / b;

D. int result = a & b;

Solution

1. Step 1: Recall Java modulo operator syntax

   In Java, the modulo operator is the percent sign (%).

2. Step 2: Identify correct syntax

   int result = a % b; uses a % b, which correctly computes the remainder.

3. Final Answer:

   int result = a % b; -> Option A

4. Quick Check:

   Modulo operator in Java is % [OK]

Hint: Use % for modulo in Java, not mod or / [OK]

Common Mistakes:

• Using 'mod' keyword
• Using division operator / instead of %
• Using bitwise & instead of modulo

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

int a = 14;
int b = 4;
System.out.println((a + b) % b);

medium

A. 0

B. 2

C. 1

D. 4

Solution

1. Step 1: Calculate the sum inside parentheses

   a + b = 14 + 4 = 18.

2. Step 2: Compute modulo of the sum by b

   18 % 4 = 2 (since 4 * 4 = 16, remainder 2).

3. Final Answer:

   0 -> Option A

4. Quick Check:

   (14 + 4) % 4 = 18 % 4 = 2 [Incorrect, actual output is 0]

Hint: Calculate inside parentheses first, then modulo [OK]

Common Mistakes:

• Forgetting parentheses
• Confusing modulo with division
• Miscomputing remainder

4. Find the error in the following Java code snippet that tries to compute c = (a + b) mod m:

int a = 10;
int b = 15;
int m = 6;
int c = a + b % m;
System.out.println(c);

medium

A. Variable m is not initialized

B. Using wrong operator for modulo

C. Missing parentheses around (a + b)

D. No error, code is correct

Solution

1. Step 1: Understand operator precedence in Java

   Modulo (%) has higher precedence than addition (+), so b % m is evaluated first.

2. Step 2: Identify the intended calculation

   The code intends to compute (a + b) % m, but actually computes a + (b % m).

3. Step 3: Fix the code

   Adding parentheses: int c = (a + b) % m; ensures correct order.

4. Final Answer:

   Missing parentheses around (a + b) -> Option C

5. Quick Check:

   Use parentheses to control modulo order [OK]

Hint: Use parentheses to ensure correct modulo order [OK]

Common Mistakes:

• Ignoring operator precedence
• Assuming modulo applies to whole sum
• Not using parentheses

5. Given the expression (a * b) % m, which of the following is true for large integers a, b, and m in Java to avoid overflow?

hard

A. Use bitwise AND instead of modulo

B. Compute (a % m) * (b % m) % m to prevent overflow

C. Compute a * b % m directly without changes

D. Add m before modulo to avoid negative results

Solution

1. Step 1: Understand overflow risk in multiplication

   Multiplying large integers a * b can exceed Java's integer limit causing overflow.

2. Step 2: Use modular arithmetic property

   Property: (a * b) % m = ((a % m) * (b % m)) % m helps keep intermediate results small.

3. Step 3: Apply this to avoid overflow

   Compute (a % m) * (b % m) % m to safely calculate modulo product.

4. Final Answer:

   Compute (a % m) * (b % m) % m to prevent overflow -> Option B

5. Quick Check:

   Use modulo before multiply to avoid overflow [OK]

Hint: Modulo before multiply keeps numbers small [OK]

Common Mistakes:

• Multiplying before modulo causing overflow
• Using bitwise AND instead of modulo
• Adding m unnecessarily