Number System
SSC-CGL Exams
1. Introduction
The Number System forms the base of all arithmetic. Every question you solve — whether it's on ratios, percentages, or equations — starts from understanding numbers deeply. Let's explore its concepts step-by-step with clear logic and exam-smart tricks.
2. Basic Concepts of Numbers
| Type | Example | Key Point |
|---|---|---|
| Natural Numbers (N) | 1, 2, 3, 4... | Counting numbers |
| Whole Numbers (W) | 0, 1, 2, 3... | Natural numbers + 0 |
| Integers (Z) | -3, -2, -1, 0, 1... | Includes negatives |
| Rational Numbers (Q) | ½, ¾, -7 | Numbers in p/q form |
| Irrational Numbers | √2, π, √5 | Non-terminating, non-repeating |
| Real Numbers (R) | All rational + irrational | Used in most problems |
Even Numbers: Divisible by 2
Odd Numbers: Not divisible by 2
Prime Numbers: Divisible by 1 and itself only (e.g. 2, 3, 5, 7, 11...)
Composite Numbers: Have more than two factors
Co-prime Numbers: HCF = 1 (e.g. 4 and 9)
3. LCM and HCF
Key Formula
HCF × LCM = Product of two numbers
⇒ ( a × b = HCF(a,b) × LCM(a,b) )
Steps to Find HCF and LCM
Example: Find HCF and LCM of 36 and 60
Prime factorization:
36 = 2² × 3²
60 = 2² × 3 × 5
HCF = 2² × 3 = 12
LCM = 2² × 3² × 5 = 180
Check: 36 × 60 = 2160, and 12 × 180 = 2160 ✓
4. Divisibility Rules
| Number | Rule |
|---|---|
| 2 | Last digit even |
| 3 | Sum of digits divisible by 3 |
| 4 | Last two digits divisible by 4 |
| 5 | Last digit 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 8 | Last three digits divisible by 8 |
| 9 | Sum of digits divisible by 9 |
| 10 | Last digit 0 |
| 11 | (Sum of odd position digits – even position digits) divisible by 11 |
Example
Check divisibility of 4,29,472 by 8
→ Last 3 digits = 472 → divisible by 8 ✓
Hence, 4,29,472 is divisible by 8.
5. Remainders and Unit Digits
Remainder Theorem
If a number ( N = dq + r ), then when N is divided by d, remainder = r.
Shortcut for Unit Digit
Unit digit of powers depends on last digit pattern.
Example:
2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 6 → repeats every 4 powers.
So, 2⁵⁷ will have same unit digit as 2³ → 8.
6. Simplification & Approximation
BODMAS Rule
Simplification means applying BODMAS rule:
Bracket → Of → Division → Multiplication → Addition → Subtraction
Example:
( 12 + 3 × (8 - 2) ÷ 3 = 12 + 3 × 6 ÷ 3 = 12 + 6 = 18 )
Approximation Tip (for MCQs)
Round numbers smartly for quick calculation.
Example:
(199.9 + 200.1) × 5 ≈ (400 × 5)/2 = 1000
7. Surds and Indices
Rules of Indices
1. ( a^m × a^n = a^{m+n} )
2. ( \frac{a^m}{a^n} = a^{m-n} )
3. ( (a^m)^n = a^{mn} )
4. ( a^0 = 1 ), ( a^{-n} = \frac{1}{a^n} )
Surds
Surd = root form of an irrational number. E.g., √2, ³√5, √7
Rationalization: multiply numerator and denominator by the conjugate.
Example:
( \frac{1}{\sqrt{3} + 1} × \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{2} )
8. Practice Set
Q1. Find LCM and HCF of 72 and 108.
View Answer
72 = 2³ × 3²
108 = 2² × 3³
HCF = 2² × 3² = 36
LCM = 2³ × 3³ = 216
Q2. What is the remainder when 45 is divided by 7?
View Answer
45 ÷ 7 → 7 × 6 = 42 → remainder = 3
Q3. Find the unit digit of ( 7^{65} ).
View Answer
Pattern of 7: (7, 9, 3, 1) repeats every 4
65 ÷ 4 → remainder 1 → unit digit = 7
Q4. Simplify: ( 64^{2/3} )
View Answer
( 64^{2/3} = (⁶⁴^{1/3})² = 4² = 16 )
Q5. Rationalize ( \frac{1}{3 + \sqrt{5}} )
View Answer
Multiply by ( 3 - \sqrt{5} ):
( \frac{1}{3 + \sqrt{5}} × \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{3 - \sqrt{5}}{4} )
Q6. If two numbers are 20 and 30, find HCF × LCM.
View Answer
HCF = 10, LCM = 60
Product = 10 × 60 = 600 = 20 × 30 ✓
Q7. Find the remainder when ( 10^{25} ) is divided by 7.
View Answer
Cycle of ( 10^n \mod 7 ): 3, 2, 6, 4, 5, 1 → repeats every 6.
25 ÷ 6 → remainder 1 → remainder = 3
Well Done!
You've completed Article 1 of Quantitative Aptitude: Number System.
Tip: Review your mistakes weekly and note down new shortcuts you discover. Mastering numbers builds 40% of your SSC CGL arithmetic confidence!
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