Forest Officer – Quantitative Aptitude
Topic: Number System
Objective – Understand the fundamentals of number systems, how to convert between them, and how to use them to solve typical aptitude problems.
1. Simple Explanation
A number system is a way of writing numbers using a set of symbols (digits) and a rule that tells how these symbols combine to represent any quantity.
| System | Base | Digits | Example (decimal 45) |
|---|---|---|---|
| Decimal | 10 | 0–9 | 45 |
| Binary | 2 | 0,1 | 101101 |
| Octal | 8 | 0–7 | 55 |
| Hexadecimal | 16 | 0–9, A–F | 2D |
Key Idea
The base (or radix) is the number of unique symbols used. Each digit’s place value is a power of the base.
2. Key Points
| Concept | What to Remember |
|---|---|
| Base‑$n$ representation | A number in base‑$n$ is written as $d_k d_{k-1}\ldots d_1 d_0$ where each $d_i$ is a digit $0\le d_i < n$. |
| Place value | Value of digit $d_i$ = $d_i \times n^i$ (rightmost digit is $i=0$). |
| Conversion (decimal → base‑$n$) | Repeatedly divide by $n$, keep remainders. Write remainders in reverse order. |
| Conversion (base‑$n$ → decimal) | Sum $d_i \times n^i$ for all digits. |
| Binary to Octal / Hex | Group binary digits in 3 (for octal) or 4 (for hex) from right to left; convert each group. |
| Common Tricks |
- Powers of 2: 2⁰=1, 2¹=2, 2²=4, 2³=8, … useful for binary‑decimal conversion.
- Powers of 8: 8⁰=1, 8¹=8, 8²=64, … useful for octal‑decimal conversion.
- Powers of 16: 16⁰=1, 16¹=16, 16²=256, … useful for hex‑decimal conversion. | | Useful Properties |
- Even / Odd in binary: last bit 0 → even, 1 → odd.
- Divisibility by 3 in base‑10: sum of digits.
- Divisibility by 7: use 2×last digit subtracted from rest. | | Common Mistakes |
- Forgetting to reverse remainders when converting to a lower base.
- Using the wrong power of the base in place‑value calculations. |
3. Worked Example
Problem
Convert the decimal number 156 to:
- Binary
- Octal
- Hexadecimal
Step‑by‑Step
1. Decimal → Binary (Base‑2)
| Division | Quotient | Remainder |
|---|---|---|
| 156 ÷ 2 | 78 | 0 |
| 78 ÷ 2 | 39 | 0 |
| 39 ÷ 2 | 19 | 1 |
| 19 ÷ 2 | 9 | 1 |
| 9 ÷ 2 | 4 | 1 |
| 4 ÷ 2 | 2 | 0 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 |
Binary: read remainders from bottom to top → 10011100₂
2. Decimal → Octal (Base‑8)
| Division | Quotient | Remainder |
|---|---|---|
| 156 ÷ 8 | 19 | 4 |
| 19 ÷ 8 | 2 | 3 |
| 2 ÷ 8 | 0 | 2 |
Octal: read remainders from bottom to top → 234₈
3. Decimal → Hexadecimal (Base‑16)
| Division | Quotient | Remainder |
|---|---|---|
| 156 ÷ 16 | 9 | 12 (C) |
| 9 ÷ 16 | 0 | 9 |
Hex: read remainders from bottom to top → 9C₁₆
4. Quick Revision Summary
| What | How to Do It |
|---|---|
| Decimal → Any Base | Divide by the base, keep remainders, reverse list. |
| Any Base → Decimal | Multiply each digit by baseⁱ and add. |
| Binary → Octal | Group bits in 3’s, convert each group. |
| Binary → Hex | Group bits in 4’s, convert each group. |
| Base‑n → Base‑m | 1) Convert to decimal 2) Convert decimal to target base. (Shortcut: binary ↔ octal/hex via grouping.) |
| Check Even/Odd | Look at last binary digit. |
| Quick Divisibility | 2/5 (last digit), 3 (digit sum), 9 (digit sum), 11 (alternating sum), 7 (2×last digit trick). |
Final Tip
When solving aptitude questions, start from the easiest base (often binary) and use grouping tricks. Always double‑check by converting back to decimal to confirm the result.