Number System – Advanced Study Notes
Target Exam: IBPS Clerk – Quantitative Aptitude
Level: MIT‑style, first‑principles with rigorous reasoning & edge‑case analysis
Why these notes?
The “Number System” section of IBPS is often the crux of quantitative aptitude. A solid grasp of base‑representation, divisibility, and modular arithmetic not only scores well in the exam but also lays a foundation for higher mathematics.
Structure
- Foundations – base‑b representation, conversions, uniqueness.
- Rational/irrational expansions – finite vs repeating, period theory.
- Divisibility – classical rules and generalizations.
- Prime‑number theory – Euclid, Wilson, Fermat.
- GCD, LCM, modular arithmetic, Euler’s theorem, CRT.
- Edge cases & pitfalls.
- Advanced topics & applications.
- Practice problems (simple → advanced).
- One challenging problem with full solution.
Notation
b – any integer base (b ≥ 2).
n – positive integer.
gcd(a,b) – greatest common divisor.
φ(n) – Euler’s totient function.
ord_b(n) – multiplicative order of b modulo n (smallest k > 0 s.t. b^k ≡ 1 (mod n)).
1. Foundations – Base‑b Representation
1.1 Decimal, Binary, Octal, Hexadecimal
- Decimal: base 10, digits 0–9.
- Binary: base 2, digits 0,1.
- Octal: base 8, digits 0–7.
- Hexadecimal: base 16, digits 0–9, A–F.
1.2 Representation in Base b
Definition
Any integer N (positive or negative) can be written uniquely as
[ N = \sum_{i=0}^{k} d_i,b^i,\quad d_i\in{0,1,\dots,b-1}, ] where k = ⌊log_b |N|⌋ for N > 0.
For negative numbers, use a minus sign or two’s complement in binary contexts.
Theorem (Uniqueness)
For a fixed base b and fixed digit set, the representation above is unique.
Proof Sketch
Suppose two distinct expansions exist. Subtracting yields a non‑zero integer that is divisible by b but less than b, impossible unless the difference is zero. ∎
1.3 Converting Between Bases
| Step | Explanation | Example (decimal → binary) |
|---|---|---|
| 1 | Divide N by b, record remainder. | 13 ÷ 2 = 6 R 1 |
| 2 | Repeat with quotient until it becomes 0. | 6 ÷ 2 = 3 R 0; 3 ÷ 2 = 1 R 1; 1 ÷ 2 = 0 R |