Modular Arithmetic Basics in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Modular Addition

What is the output of the following JavaScript code that performs modular addition?

DSA Javascript

const a = 17;
const b = 23;
const mod = 10;
const result = (a + b) % mod;
console.log(result);

A 9

B 10

C 1

D 0

💡 Hint

Remember to add first, then apply the modulo operation.

The sum of 17 and 23 is 40. Taking 40 modulo 10 gives the remainder 0, so the correct output is 0.

❓ Predict Output

intermediate

Modular Multiplication Result

What will be printed after running this JavaScript code that calculates modular multiplication?

DSA Javascript

const x = 7;
const y = 8;
const mod = 5;
const product = (x * y) % mod;
console.log(product);

A 1

B 6

C 0

D 2

💡 Hint

Multiply first, then take modulo.

7 times 8 is 56. 56 modulo 5 is 1 because 5 * 11 = 55 and remainder is 1.

❓ Predict Output

advanced

Modular Exponentiation Output

What is the output of this code that calculates modular exponentiation using repeated multiplication?

DSA Javascript

let base = 3;
let exponent = 4;
let mod = 7;
let result = 1;
for(let i = 0; i < exponent; i++) {
  result = (result * base) % mod;
}
console.log(result);

A 2

B 4

C 5

D 1

💡 Hint

Calculate 3^4 mod 7 step by step.

3^4 = 81. 81 modulo 7 is 4 because 7 * 11 = 77 and remainder is 4.

🧠 Conceptual

advanced

Properties of Modular Arithmetic

Which of the following statements about modular arithmetic is TRUE?

A Modular addition is not associative.

B Modular subtraction always produces a positive result.

C Modular multiplication distributes over modular addition.

D Modular division is always defined for any divisor.

💡 Hint

Think about how multiplication and addition behave under modulo.

Modular multiplication distributes over modular addition: (a * (b + c)) % m = ((a * b) % m + (a * c) % m) % m.

Modular addition is associative, so A is false. Modular subtraction can be negative before modulo, so C is false. Modular division is not always defined, so D is false.

🔧 Debug

expert

Identify the Error in Modular Inverse Calculation

Consider this code snippet that attempts to find the modular inverse of a number. What error will it produce when run?

DSA Javascript

function modInverse(a, m) {
  for(let x = 1; x < m; x++) {
    if ((a * x) % m == 1) {
      return x;
    }
  }
  return null;
}
console.log(modInverse(6, 9));

A Returns null

B Infinite loop

C Throws a TypeError

D Returns 4

💡 Hint

Check if 6 has an inverse modulo 9.

6 and 9 are not coprime, so modular inverse does not exist. The function returns null.

Practice

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

easy

A. 2

B. 3

C. 1

D. 0

Solution

1. Step 1: Understand the modulus operation

   The modulus operator (%) gives the remainder after division of the first number by the second.

2. Step 2: Calculate 7 divided by 3

   7 divided by 3 is 2 with a remainder of 1.

3. Final Answer:

   1 -> Option C

4. Quick Check:

   7 % 3 = 1 [OK]

Hint: Modulus gives remainder after division [OK]

Common Mistakes:

• Confusing modulus with division
• Using 3 instead of remainder
• Thinking modulus returns quotient

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

easy

A. a % b

B. a mod b

C. mod(a, b)

D. a // b

Solution

1. Step 1: Recall JavaScript modulus syntax

   In JavaScript, the modulus operator is the percent sign (%) placed between two numbers or variables.

2. Step 2: Identify correct syntax

   Only a % b correctly computes the remainder of a divided by b.

3. Final Answer:

   a % b -> Option A

4. Quick Check:

   Use % for modulus in JavaScript [OK]

Hint: Use % operator for modulus in JavaScript [OK]

Common Mistakes:

• Using 'mod' keyword
• Using integer division operator //
• Calling a non-existent mod() function

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

const x = 15;
const y = 4;
console.log(x % y);

medium

A. 4

B. 3

C. 0

D. 1

Solution

1. Step 1: Calculate 15 divided by 4

   15 divided by 4 is 3 with a remainder of 3.

2. Step 2: Apply modulus operator

   The expression 15 % 4 returns the remainder, which is 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   15 % 4 = 3 [OK]

Hint: Modulus returns remainder after division [OK]

Common Mistakes:

• Confusing remainder with quotient
• Expecting 4 instead of remainder
• Mistaking modulus for division

4. Find the error in the following code snippet that tries to compute the modulus of two numbers:

let a = 10;
let b = 3;
let result = a % b();
console.log(result);

medium

A. Missing semicolon after console.log

B. Modulus operator should be replaced with division

C. Variables a and b should be strings

D. Using parentheses after b causes an error

Solution

1. Step 1: Analyze the modulus expression

   The code uses a % b(), which treats b as a function call, but b is a number, not a function.

2. Step 2: Identify the error

   Parentheses after b cause a runtime error because numbers cannot be called like functions.

3. Final Answer:

   Using parentheses after b causes an error -> Option D

4. Quick Check:

   Variables are not functions, so no () after b [OK]

Hint: Don't add () after variables unless they are functions [OK]

Common Mistakes:

• Adding () after number variables
• Confusing modulus with division
• Ignoring runtime errors

5. You want to find the remainder when the sum of two numbers a = 123456789 and b = 987654321 is divided by 1000. Which expression correctly computes this using modular arithmetic properties?

hard

A. ((a % 1000) + (b % 1000)) % 1000

B. (a % 1000) + (b % 1000)

C. a + b % 1000

D. a % (b % 1000)

Solution

1. Step 1: Recall modular addition property

   The remainder of a sum modulo m equals the sum of remainders modulo m: (a + b) % m = ((a % m) + (b % m)) % m.

2. Step 2: Apply property to given numbers

   Compute (a % 1000) and (b % 1000), add them, then take modulo 1000 again to get correct remainder.

3. Final Answer:

   ((a % 1000) + (b % 1000)) % 1000 -> Option A

4. Quick Check:

   Use modular addition: ((a%m)+(b%m))%m [OK]

Hint: Add remainders then mod again for sum modulus [OK]

Common Mistakes:

• Skipping final modulus after addition
• Adding without modulus
• Using wrong modulus order