Overview
Number System is one of the most fundamental and high-weightage topics in the **SSC CGL** Quantitative Aptitude section. In Tier 1, the Quantitative Aptitude section carries **50 marks (25 questions)**, and Number System concepts directly or indirectly contribute to **3-5 questions**. Many other topics like Percentage, Profit & Loss, and Time & Work build upon Number System fundamentals.
For SSC CGL, you need a strong grasp of divisibility rules, HCF/LCM, remainders, and properties of numbers. Questions range from direct concept-based to tricky application-based problems.
Key Concepts
### Classification of Numbers
1. **Natural Numbers (N)**: 1, 2, 3, 4, 5... (positive counting numbers)
2. **Whole Numbers (W)**: 0, 1, 2, 3, 4... (natural numbers + zero)
3. **Integers (Z)**: ...-3, -2, -1, 0, 1, 2, 3... (positive, negative, and zero)
4. **Rational Numbers (Q)**: Numbers expressible as p/q where q ≠ 0 (e.g., 1/2, 3/4, 0.75)
5. **Irrational Numbers**: Cannot be expressed as p/q (e.g., √2, √3, π)
6. **Real Numbers (R)**: All rational and irrational numbers combined
7. **Prime Numbers**: Divisible only by 1 and itself (2, 3, 5, 7, 11, 13...)
8. **Composite Numbers**: Have more than two factors (4, 6, 8, 9, 10...)
### Important: 1 is neither prime nor composite. 2 is the only even prime number.
Detailed Explanation
### Divisibility Rules (Critical for SSC CGL)
| Divisor | Rule | Example |
|---|---|---|
| **2** | Last digit is even (0,2,4,6,8) | 246 → last digit 6 (even) ✓ |
| **3** | Sum of digits divisible by 3 | 123 → 1+2+3=6 → 6÷3=2 ✓ |
| **4** | Last 2 digits divisible by 4 | 1324 → 24÷4=6 ✓ |
| **5** | Last digit is 0 or 5 | 345 → last digit 5 ✓ |
| **6** | Divisible by both 2 AND 3 | 312 → even ✓, 3+1+2=6÷3=2 ✓ |
| **7** | Double last digit, subtract from rest | 343 → 34-(3×2)=34-6=28 → 28÷7=4 ✓ |
| **8** | Last 3 digits divisible by 8 | 1320 → 320÷8=40 ✓ |
| **9** | Sum of digits divisible by 9 | 729 → 7+2+9=18 → 18÷9=2 ✓ |
| **11** | |Sum of odd-placed digits - Sum of even-placed digits| divisible by 11 | 1023 → (1+2)-(0+3)=0 → 0÷11=0 ✓ |
### HCF (Highest Common Factor) and LCM (Least Common Multiple)
**HCF Methods:**
1. **Prime Factorization**: Find common prime factors, take lowest power
2. **Division Method**: Divide larger by smaller, continue with remainder
**LCM Methods:**
1. **Prime Factorization**: Take all prime factors with highest power
2. **Division Method**: Divide by smallest prime, continue until all become 1
**Key Formula**: HCF × LCM = Product of two numbers
- For numbers a and b: HCF(a,b) × LCM(a,b) = a × b
**Example**: Find HCF and LCM of 12 and 18.
- 12 = 2² × 3, 18 = 2 × 3²
- HCF = 2¹ × 3¹ = 6
- LCM = 2² × 3² = 36
- Verify: 6 × 36 = 216 = 12 × 18 ✓
### Remainder Theorem
**Basic Formula**: Dividend = Divisor × Quotient + Remainder
**Key Properties:**
- Remainder of (a + b) ÷ n = (Rem(a÷n) + Rem(b÷n)) mod n
- Remainder of (a × b) ÷ n = (Rem(a÷n) × Rem(b÷n)) mod n
- Remainder of (aⁿ) ÷ d: Use cyclicity
### Cyclicity of Remainders
The last digit of powers follows a cycle:
| Digit | Cycle | Period |
|---|---|---|
| 0 | 0,0,0,0 | 1 |
| 1 | 1,1,1,1 | 1 |
| 2 | 2,4,8,6 | 4 |
| 3 | 3,9,7,1 | 4 |
| 4 | 4,6,4,6 | 2 |
| 5 | 5,5,5,5 | 1 |
| 6 | 6,6,6,6 | 1 |
| 7 | 7,9,3,1 | 4 |
| 8 | 8,4,2,6 | 4 |
| 9 | 9,1,9,1 | 2 |
**Example**: Find the last digit of 7²⁵.
- Cycle of 7: 7, 9, 3, 1 (period 4)
- 25 ÷ 4 = 6 remainder 1
- Last digit = 1st in cycle = **7**
### Counting Prime Numbers
**Primes up to 100** (25 primes): 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
**To check if n is prime**: Check divisibility by all primes up to √n.
- To check if 97 is prime: √97 ≈ 9.8, check primes 2,3,5,7 → none divide 97 → prime ✓