📚 Punjab Patwari / Lekhpal – Quantitative Aptitude
Topic: Number System
Objective – Understand the different numeral systems, how to convert between them, and key properties that help solve exam‑type questions quickly.
1️⃣ Simple Explanation
| Concept | What it is | Example |
|---|---|---|
| Decimal (Base‑10) | The standard system we use daily. | 243₁₀ |
| Binary (Base‑2) | Uses only 0 and 1. | 111101₁₀ = 61₁₀ |
| Octal (Base‑8) | Uses digits 0‑7. | 345₈ = 221₁₀ |
| Hexadecimal (Base‑16) | Uses 0‑9 and A‑F. | 1A3₁₆ = 419₁₀ |
Place‑Value – In any base b, the rightmost digit has value b⁰, next b¹, then b², etc.
Example: In 3A₁₆, 3×16¹ + A×16⁰ = 3×16 + 10×1 = 58₁₀.
2️⃣ Key Points to Remember
| # | Property / Rule | Why it matters |
|---|---|---|
| 1. Base Conversion | Use repeated division or multiplication by the base. | Essential for any conversion question. |
| 2. Divisibility Rules | - Binary: last digit 0 → even, 1 → odd. - Octal: last digit even → divisible by 2; last two digits form a number divisible by 4 → divisible by 4. - Hex: last digit 0–7 → divisible by 2; last digit 0,4,8,C → divisible by 4. |
Quickly check divisibility without full conversion. |
| 3. Number of Digits | ⌊log_b N⌋ + 1 | Helps estimate size & check for overflow. |
| 4. Casting Out Nines | Sum of decimal digits ≡ number (mod 9). | Quick checksum for decimal numbers. |
| 5. Repeating Fractions | Binary fraction 0.1011₁₀ = 0.1011₂ (finite) ; 0.101₁₀ = 0.001100110011…₂ (repeating). | Important for problems involving fractions in binary. |
| 6. Powers of 2, 8, 16 | 2⁰=1, 2¹=2 …; 8⁰=1, 8¹=8 …; 16⁰=1, 16¹=16 … | Needed when checking if a number is a power or multiple of these. |
| 7. Conversion Tables | Keep a quick reference for 0‑15 in binary, octal, hex. | Saves time during exam. |
3️⃣ Worked Example
Problem
Convert the following decimal number 156 into binary, octal, and hexadecimal.
Also, convert the binary number 101101₂ to decimal.
Part A – Decimal 156 ➜ Binary, Octal, Hex
| Step | Binary | Octal | Hexadecimal |
|---|---|---|---|
| 1️⃣ | Divide 156 by 2 → quotient 78, remainder 0 (LSB). | Divide 156 by 8 → quotient 19, remainder 4 (LSB). | Divide 156 by 16 → quotient 9, remainder 12 (C). |
| 2️⃣ | 78 ÷ 2 = 39 r 0 | 19 ÷ 8 = 2 r 3 | 9 ÷ 16 = 0 r 9 |
| 3️⃣ | 39 ÷ 2 = 19 r 1 | 2 ÷ 8 = 0 r 2 | 0 → stop |
| 4️⃣ | 19 ÷ 2 = 9 r 1 | – | – |
| 5️⃣ | 9 ÷ 2 = 4 r 1 | – | – |
| 6️⃣ | 4 ÷ 2 = 2 r 0 | – | – |
| 7️⃣ | 2 ÷ 2 = 1 r 0 | – | – |
| 8️⃣ | 1 ÷ 2 = 0 r 1 | – | – |
Now read remainders bottom‑up:
- Binary: 10011100₂
- Octal: 234₈
- Hex: 9C₁₆
Check:
1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰ = 128+0+0+16+8