Number System
Basics of numbers, place value, divisibility, HCF/LCM, remainders, and base conversions.
What is the Number System?
A number system is a way to represent and work with quantities. Competitive exams test definitions, shortcuts, and properties—especially divisibility, remainders (mod), HCF/LCM, and place value tricks.
Key sets & terms
- ℕ: Natural (1,2,3,…)
- 𝕎: Whole (0,1,2,…)
- ℤ: Integers (…−2,−1,0,1,…)
- ℚ: Rationals (p/q, q≠0)
- ℝ: Reals (rationals + irrationals)
- Prime: >1 with exactly two factors
Place Value & Representation
In base-10, each digit’s value depends on its position (×10, ×100, …).
| Number | Expanded Form | Notes |
|---|---|---|
| 5,708 | 5×10³ + 7×10² + 0×10¹ + 8×10⁰ | Zero can hold a place but adds no value. |
| 203.45 | 2×10² + 0×10¹ + 3×10⁰ + 4×10⁻¹ + 5×10⁻² | Right of decimal → negative powers of 10. |
| n = 10a + b | a = tens digit, b = units digit | Useful in mod 9 / digit-sum problems. |
⌊N/5⌋ + ⌊N/25⌋ + ⌊N/125⌋ + …
Common Divisibility Rules
| k | Rule (base-10) | Example |
|---|---|---|
| 2 | Last digit even | 748 ✓ |
| 3 | Sum of digits divisible by 3 | 7+4+9=20 → no |
| 4 | Last two digits divisible by 4 | 312 → yes |
| 5 | Ends with 0 or 5 | 985 ✓ |
| 8 | Last three digits divisible by 8 | 1024 → yes |
| 9 | Digit sum divisible by 9 | 3+6+9=18 → yes |
| 11 | Alt. sum of digits divisible by 11 | (1−2+1)=0 → yes for 121 |
Use prime factorization for combined checks (e.g., 12 = 3×4).
HCF/LCM Essentials
- HCF (GCD): greatest common divisor
- LCM: least common multiple
- Product relation:
a×b = HCF(a,b) × LCM(a,b)(for positive integers)
- 72 = 2³×3²
- 108 = 2²×3³
- HCF = 2²×3² = 36
- LCM = 2³×3³ = 216
// Euclid GCD (pseudo-code)
gcd(a, b):
while b ≠ 0:
(a, b) = (b, a mod b)
return aLCM(a,b)=a×b and HCF(a,b)=1.
Remainders & Congruence (mod)
- a ≡ b (mod m) means m divides (a−b).
- Work with remainders: reduce early to keep numbers small.
- (a+b) mod m = ((a mod m)+(b mod m)) mod m
- (a·b) mod m = ((a mod m)·(b mod m)) mod m
- a<sup>k</sup> mod m often uses patterns/cycles.
Base-n Conversion (Quick)
Check digits: each d<sub>i</sub> must be < base.
Practice
A) Multiple Choice
-
The units digit of 3<sup>57</sup> is:
1 3 7 9
-
HCF(84, 126) equals:
14 21 42 63
-
The number divisible by 11 is:
1,232 1,254 1,331 1,421
B) Fill in the Blanks
- Trailing zeros in 100! = _______. (use powers of 5)
- If a ≡ 7 (mod 9) and b ≡ 5 (mod 9), then (a+b) ≡ _______ (mod 9).
- LCM(18, 24) = _______. (use prime factors)
- (10101)<sub>2</sub> in decimal = _______.
C) Tiny code sample (formatting test)
// Fast digit cycle for last digit (mod 10)
const lastDigit = (a, n) => {
const cycle = [a%10];
while (true) {
const next = (cycle.at(-1) * (a%10)) % 10;
if (next === cycle[0]) break;
cycle.push(next);
}
return cycle[(n-1) % cycle.length];
};Short Reading
A factory produces items in batches of 18 and packs them into boxes of 24. To combine batches with no leftovers, they must produce a count that is a multiple of both—i.e., the LCM. For quick checks, use prime factorization: 18 = 2×3² and 24 = 2³×3 ⇒ LCM = 2³×3² = 72.
Task: If each item weighs 250 g, how many kilograms are in the minimum combined shipment?
Show Suggested Answers
MCQ
- 7 (cycle of 3: 3,9,7,1; 57 mod 4 = 1 → 3; wait, for 3 the cycle is 3,9,7,1 → 57 mod 4 = 1 → 3)
- 42 (84=2²·3·7; 126=2·3²·7 → HCF=2·3·7)
- 1,331 (1−3+3−1=0 → divisible by 11)
Fill in the Blanks
- 24 (⌊100/5⌋=20, ⌊100/25⌋=4; total 24)
- 3 (7+5=12 ≡ 3 mod 9)
- 72 (18=2·3², 24=2³·3 → LCM=2³·3²)
- 21 (16+0+4+0+1)
Reading
Minimum shipment = LCM count = 72 items; weight = 72×0.25 kg = 18 kg.
Exam tips
- Reduce numbers early with mod.
- Use prime powers for HCF/LCM quickly.
- Memorize small digit cycles (2,3,4,7,8,9).
- Check divisibility before long division.
