How to Solve: CUET Quant – Number System (LCM, HCF, Divisibility, Remainder Theorem)

How to Solve: CUET Quant – Number System (LCM, HCF, Divisibility, Remainder Theorem)

Introduction

"If you can find the LCM of 12 and 18 in 10 seconds, you just saved 2 minutes on your CUET exam—and those 2 minutes could be the difference between a 90 and a 99. Let’s make sure you never lose marks on Number Systems again."

What You Need To Know First

1. Prime Factorization – Breaking numbers into products of primes (e.g., 12 = 2² × 3).
2. Basic Division – How to divide numbers and interpret remainders.
3. Multiples & Factors – What it means for a number to be a multiple or factor of another.

(If you’re shaky on any of these, pause and review them first—this guide assumes you’re solid.)

Key Vocabulary

Formulas To Know

1. LCM of Two Numbers (Prime Factorization Method)

Formula: LCM(a, b) = (a × b) / HCF(a, b) OR LCM(a, b) = Product of the highest powers of all primes in a and b

Variables: - a, b = Two numbers - HCF(a, b) = Highest Common Factor of a and b

MEMORISE THIS – Not always given on the exam sheet.

2. HCF of Two Numbers (Prime Factorization Method)

Formula: HCF(a, b) = Product of the lowest powers of common primes in a and b

MEMORISE THIS – Essential for LCM and problem-solving.

3. Relationship Between LCM and HCF

Formula: LCM(a, b) × HCF(a, b) = a × b

MEMORISE THIS – Saves time in word problems.

4. Divisibility Rules (Quick Checks)

MEMORISE THIS – These rules appear in every CUET exam.

5. Remainder Theorem (Euclid’s Division Lemma)

Formula: Dividend = (Divisor × Quotient) + Remainder OR a = bq + r, where 0 ≤ r < b

Variables: - a = Dividend (number being divided) - b = Divisor - q = Quotient - r = Remainder

Given on exam sheet? Sometimes, but MEMORISE IT—it’s the foundation of remainder problems.

Step-by-Step Method

How to Find LCM (Prime Factorization Method)

Steps: 1. Prime factorize both numbers. 2. List all primes that appear in either number. 3. Take the highest power of each prime. 4. Multiply them together to get LCM.

Example: Find LCM of 12 and 18. 1. 12 = 2² × 3¹
18 = 2¹ × 3² 2. Primes: 2, 3 3. Highest powers: 2², 3² 4. LCM = 2² × 3² = 4 × 9 = 36

How to Find HCF (Prime Factorization Method)

Steps: 1. Prime factorize both numbers. 2. List common primes in both. 3. Take the lowest power of each common prime. 4. Multiply them together to get HCF.

Example: Find HCF of 12 and 18. 1. 12 = 2² × 3¹
18 = 2¹ × 3² 2. Common primes: 2, 3 3. Lowest powers: 2¹, 3¹ 4. HCF = 2¹ × 3¹ = 2 × 3 = 6

How to Solve Remainder Problems (Using Remainder Theorem)

Steps: 1. Write the equation: Dividend = (Divisor × Quotient) + Remainder 2. Plug in known values (usually remainder and divisor are given). 3. Solve for the unknown (often the dividend or quotient). 4. Check constraints: Remainder must be less than divisor.

Example: A number when divided by 5 leaves a remainder 3. If the quotient is 4, find the number. 1. Dividend = (5 × 4) + 3 2. Dividend = 20 + 3 = 23 3. Answer: 23

Worked Examples

Example 1 – Basic LCM & HCF

Question: Find LCM and HCF of 24 and 36.

Solution: 1. Prime factorize:
- 24 = 2³ × 3¹
- 36 = 2² × 3² 2. LCM: Highest powers → 2³ × 3² = 8 × 9 = 72 3. HCF: Lowest powers → 2² × 3¹ = 4 × 3 = 12

What we did and why: - Broke numbers into primes to systematically find LCM/HCF. - Used highest powers for LCM, lowest for HCF—this is the only reliable method.

Example 2 – Medium (Word Problem)

Question: Two bells ring every 18 and 24 seconds respectively. If they ring together at 12:00 PM, when will they ring together again?

Solution: 1. Find LCM of 18 and 24 (time interval when both ring together).
- 18 = 2 × 3²
- 24 = 2³ × 3
- LCM = 2³ × 3² = 8 × 9 = 72 seconds 2. Convert 72 seconds to minutes: 72 ÷ 60 = 1 minute 12 seconds. 3. Next ring time: 12:00 PM + 1 min 12 sec = 12:01:12 PM

What we did and why: - Recognized that LCM gives the next common time for repeating events. - Converted seconds to minutes for real-world context (exam trick).

Example 3 – Exam Style (Remainder + Divisibility)

Question: A number when divided by 7 leaves a remainder 4. What is the remainder when 3 times the number is divided by 7?

Solution: 1. Let the number = N. 2. Given: N = 7q + 4 (Remainder Theorem). 3. Multiply by 3: 3N = 3(7q + 4) = 21q + 12. 4. Divide 3N by 7: (21q + 12) ÷ 7 = 3q + (12 ÷ 7). 5. 12 ÷ 7 leaves remainder 5 (since 7 × 1 = 7, 12 - 7 = 5). 6. Final remainder = 5.

What we did and why: - Used Remainder Theorem to express the number algebraically. - Multiplied remainder (4 × 3 = 12) and re-divided by 7 to find new remainder. - Key insight: Remainders multiply when the number is scaled.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, CUET warriors—here’s your 60-second crash recap for Number Systems:

1. LCM = Highest powers of all primes in the numbers. HCF = Lowest powers of common primes.
2. LCM × HCF = Product of the two numbers—use this to check your answers.
3. Divisibility rules are your best friends—memorize them cold.
4. Remainder Theorem: Dividend = Divisor × Quotient + Remainder. If you multiply the number, multiply the remainder too.
5. Exam traps? Watch for word problems hiding LCM/HCF and negative remainders.

Tonight, do 5 problems—2 LCM/HCF, 2 divisibility, 1 remainder. Tomorrow, you’ll solve these in your sleep. Good luck—you’ve got this!