Number System (Advanced)
Learning Goals
By the end of this chapter, you will be able to:
- Classify numbers confidently: natural, whole, integers, rational, irrational, real, and complex.
- Apply divisibility rules and prime factorisation to solve LCM–HCF problems in seconds.
- Master remainders, unit digits, and cyclicity — the favourite hunting ground of CAT, SSC CGL Tier-II, and IBPS PO Mains.
- Handle base systems (binary, octal, hexadecimal) and conversions without fear.
- Tackle factorial-based questions (trailing zeros, highest power dividing n!) with a one-line method.
- Spot the traps examiners set: "almost-true" statements, sign confusion, and remainder-vs-divisor mix-ups.
The Hook: A Number That Broke a Computer
In 1996, the European Space Agency's Ariane 5 rocket exploded less than a minute after launch. The cause? A number. A 64-bit floating-point value was squeezed into a 16-bit integer — the number was too big for its "box," the system overflowed, and a rocket worth hundreds of millions of dollars turned into fireworks.
Numbers are not just symbols. They live in *systems*, and every system has rules. In your exam, the "explosion" is quieter — one wrong assumption about remainders or unit digits, and 2 marks vanish. This chapter teaches you the rules of the box, so your numbers never overflow.
1. The Family Tree of Numbers
Let's be honest — most aspirants *think* they know number types, then lose marks on a statement like "Every integer is a rational number." (It is true, by the way.) So let's build the tree properly.
- Natural numbers (N): 1, 2, 3, … — counting numbers.
- Whole numbers (W): 0, 1, 2, 3, … — natural numbers plus zero.
- Integers (Z): …, −2, −1, 0, 1, 2, …
- Rational numbers (Q): numbers expressible as p/q, where p, q are integers and q ≠ 0. Decimals that terminate (0.25) or repeat (0.333…) are rational.
- Irrational numbers: cannot be written as p/q — √2, √3, π, e. Their decimals never terminate and never repeat.
- Real numbers (R): rational + irrational — everything on the number line.
- Complex numbers: a + bi, where i = √−1. (Rarely asked, but know the definition.)
Key exam facts:
- 0 is a whole number but *not* a natural number.
- Every integer is rational (5 = 5/1), but not every rational is an integer (3/4).
- The sum of a rational and an irrational number is always irrational. The product of two irrationals *can* be rational (√2 × √2 = 2) — a classic trap!
🧠 Active Recall: Close the book. Is 0.999… (repeating forever) rational or irrational? Why? *(Hint: it equals exactly 1.)*
2. Prime Numbers and Factorisation — The Atoms of Arithmetic
A prime number has exactly two factors: 1 and itself. 2 is the only even prime. 1 is *neither* prime nor composite — examiners love this.
Primes up to 100 (25 of them): 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.
Fundamental Theorem of Arithmetic: Every integer greater than 1 can be written as a product of primes in exactly one way (order aside). Example: 360 = 2³ × 3² × 5.
From prime factorisation N = aᵖ × bᑫ × cʳ …:
- Number of factors = (p+1)(q+1)(r+1)…
- Sum of factors = [(aᵖ⁺¹−1)/(a−1)] × [(bᑫ⁺¹−1)/(b−1)] × …
- Number of even factors = p × (q+1)(r+1)… (when a = 2)
- Number of ways to write N as a product of two factors = (number of factors)/2, if N is not a perfect square; (factors+1)/2 if it is.
🧠 Active Recall: How many factors does 360 have? Compute it before reading on. *(Answer: (3+1)(2+1)(1+1) = 24.)*
3. Divisibility Rules — Your Speed Weapons
| Divisor | 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 |
| 7 | Double the last digit, subtract from the rest; result divisible by 7 |
| 8 | Last three digits divisible by 8 |
| 9 | Sum of digits divisible by 9 |
| 11 | (Sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11 |
| 12 | Divisible by both 3 and 4 |
| 25 | Last two digits 00, 25, 50, or 75 |
Composite divisors: To test divisibility by 72, test 8 and 9 (co-prime pair), *not* 6 and 12 — because 6 and 12 share factors. Always split into co-prime factors.
🧠 Active Recall: Is 4,95,432 divisible by 11? By 8? Do it mentally.
4. LCM and HCF — The Twin Pillars
- HCF (GCD): the largest number dividing all given numbers.
- LCM: the smallest number divisible by all given numbers.
Golden relation (for two numbers): LCM × HCF = Product of the two numbers. ⚠️ This works only for *two* numbers — a favourite trap.
Fractions:
- LCM of fractions = LCM of numerators / HCF of denominators
- HCF of fractions = HCF of numerators / LCM of denominators
Remainder-flavoured models (asked constantly):
- "Greatest number that divides a, b, c leaving the *same* remainder r" → HCF of (a−r, b−r, c−r).
- "Greatest number dividing a, b, c leaving remainders p, q, s" → HCF of (a−p, b−q, c−s).
- "Smallest number which when divided by a, b, c leaves remainder r each time" → LCM(a, b, c) + r.
- "Smallest number which when divided by a, b, c leaves remainders p, q, s where a−p = b−q = c−s = k" → LCM(a, b, c) − k.
🧠 Active Recall: Why does the last formula *subtract* k? Say the reason aloud in one sentence.
5. Remainders — Where Toppers Are Made
Division algorithm: Dividend = Divisor × Quotient + Remainder, with 0 ≤ Remainder < Divisor.
Negative remainders: When 47 is divided by 8, remainder is 7. But thinking of it as "−1" (since 47 = 8×6 − 1) often makes calculations faster. Negative remainder = actual remainder − divisor.
Powerful remainder facts:
- (a × b) mod m = [(a mod m) × (b mod m)] mod m
- (a + b) mod m = [(a mod m) + (b mod m)] mod m
- aⁿ mod m: find the cycle of powers, then reduce n modulo the cycle length.
Classic results:
- xⁿ − aⁿ is divisible by (x − a) for all n.
- xⁿ − aⁿ is divisible by (x + a) only when n is even.
- xⁿ + aⁿ is divisible by (x + a) only when n is odd.
- Fermat's Little Theorem: If p is prime and a is not a multiple of p, then aᵖ⁻¹ ≡ 1 (mod p).
- Wilson's Theorem: (p−1)! ≡ −1 (mod p) for prime p.
Example: Remainder when 2¹⁰⁰ is divided by 7? Powers of 2 mod 7 cycle: 2, 4, 1 (cycle length 3). 100 = 3×33 + 1, so remainder = 2.
🧠 Active Recall: Find the remainder when 3⁵⁰ is divided by 7 using Fermat's theorem. *(3⁶ ≡ 1, so 3⁴⁸ ≡ 1, 3⁵⁰ ≡ 9 ≡ 2.)*
6. Unit Digits and Cyclicity
The unit digit of aⁿ depends only on the unit digit of a, and it cycles:
| Unit digit of a | Cycle | Length |
|---|---|---|
| 0, 1, 5, 6 | same always | 1 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 4 | 4, 6 | 2 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 9 | 9, 1 | 2 |
Method: divide the exponent by the cycle length; use the remainder (if remainder is 0, use the last position of the cycle).
Example: Unit digit of 7⁹⁵? Cycle of 7 is 7, 9, 3, 1. 95 ÷ 4 leaves remainder 3 → third position → 3.
🧠 Active Recall: Unit digit of 2²⁰²⁴? *(2024 ÷ 4 remainder 0 → 4th position → 6.)*
7. Factorials — Trailing Zeros and Highest Powers
n! = 1 × 2 × 3 × … × n.
Trailing zeros in n! = number of times 5 divides n! (2s are always plentiful):
Zeros = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …
Example: Zeros in 100! = 20 + 4 = 24.
Highest power of a prime p dividing n! = ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + …
For a *composite* divisor (say, highest power of 6 = 2×3 in n!), find the powers of each prime and take the limiting one.
🧠 Active Recall: How many trailing zeros does 250! have? *(50 + 10 + 2 = 62.)*
8. Base Systems — Thinking Like a Computer
We count in base 10. Computers use base 2 (binary), base 8 (octal), base 16 (hexadecimal: digits 0–9, A–F).
Base b → decimal: multiply each digit by b raised to its position (from right, starting at 0).
(1011)₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11.
Decimal → base b: divide repeatedly by b, collect remainders bottom-to-top.
Binary ↔ Octal/Hex shortcut: group binary digits in 3s (octal) or 4s (hex) from the right.
(101101)₂ → groups 101 | 101 → (55)₈.
🧠 Active Recall: Convert (45)₁₀ to binary. *(101101)*
9. The Mnemonic Corner 🪄
"2-4-8-6, the Even Fix" — the cycle of 2 is all even digits in order: 2, 4, 8, 6. And 8's cycle is the *reverse*: 8, 4, 2, 6. Remember: "2 climbs up, 8 climbs down."
For divisibility by 11: "Odd minus Even, Zero or Eleven" — alternate digit sums must differ by 0 or a multiple of 11.
10. Examiner's Traps — Read Before You Regret
- "Every whole number is a natural number" — FALSE (0 is whole, not natural).
- "Product of two irrationals is always irrational" — FALSE (√2 × √8 = 4).
- LCM × HCF = product — only for *two* numbers.
- Remainder must be less than the divisor — options often include a value equal to the divisor.
- In "leaves remainder r" problems, check whether the question asks for the *smallest* or *greatest* such number.
- Cyclicity: when the exponent divides exactly (remainder 0), take the last element of the cycle, not the first.
Summary Table
| Concept | Core Result |
|---|---|
| Factors of N = aᵖbᑫcʳ | (p+1)(q+1)(r+1) |
| LCM × HCF | = product (two numbers only) |
| Same remainder r on dividing a, b, c | HCF of (a−r, b−r, c−r) |
| Smallest number leaving remainder r with divisors a, b, c | LCM(a,b,c) + r |
| Unit digit of aⁿ | Reduce n mod cycle length (0 → last position) |
| Trailing zeros in n! | ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + … |
| Highest power of prime p in n! | ⌊n/p⌋ + ⌊n/p²⌋ + … |
| Fermat's Little Theorem | aᵖ⁻¹ ≡ 1 (mod p), p prime, p ∤ a |
| Divisibility by 11 | Alternate-sum difference = 0 or multiple of 11 |
| Binary → decimal | Σ digit × 2^position |
🔁 Review This After 2 Days
Don't just re-read. Cover the summary table and answer aloud:
- How many factors does 720 have?
- Remainder when 7⁸⁴ is divided by 5?
- Trailing zeros in 200!?
- Convert (110101)₂ to decimal.
- Smallest number which when divided by 4, 5, 6 leaves remainder 3 each time?
If any answer takes more than 30 seconds, revisit that section — then test yourself again after another 2 days. Spaced repetition is how this chapter moves from "studied" to "scored."