Number System

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

The Number System is the foundation of Quantitative Aptitude for SSC CGL. It covers natural numbers, whole numbers, integers, rational numbers, and their properties. Questions test your understanding of divisibility rules, factors, multiples, HCF and LCM, remainders, and patterns in sequences.

Types of Numbers:

Type	Definition	Examples
Natural numbers	Counting numbers (1, 2, 3…)	1, 2, 3…
Whole numbers	Natural numbers + 0	0, 1, 2, 3…
Integers	Whole numbers + negatives	…-3, -2, -1, 0, 1, 2…
Rational numbers	p/q form where q≠0	1/2, -3/4, 5
Irrational numbers	Cannot be expressed as p/q	√2, π, e
Real numbers	Rational + Irrational	All numbers on number line

Divisibility Rules:

Divisor	Rule
2	Last digit even (0, 2, 4, 6, 8)
3	Sum of digits divisible by 3
4	Last two digits divisible by 4
5	Last digit 0 or 5
6	Divisible by 2 AND 3
8	Last three digits divisible by 8
9	Sum of digits divisible by 9
11	(Sum of odd-position digits) - (Sum of even-position digits) divisible by 11

⚡ Exam Tip: For 11, example: 3521 → (3+1) - (5+2) = 4 - 7 = -3, not divisible by 11. 3524 → (3+4) - (5+2) = 7 - 7 = 0, divisible by 11.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and problem-solving practice.

Factors and Multiples:

• Factor (Divisor): A number that divides another exactly (e.g., 3 is factor of 12)
• Multiple: A number that is divisible by another (e.g., 12 is multiple of 3)
• Prime number: Divisible only by 1 and itself (2, 3, 5, 7, 11, 13…)
• Composite: Has more than two factors
• Co-prime (Relatively prime): Two numbers with HCF = 1

Prime Numbers: First 20 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71

Note: 1 is neither prime nor composite. 2 is the only even prime.

HCF (Highest Common Factor / GCD):

Method 1: Prime factorisation Find prime factors of each number; HCF = product of common prime factors with minimum powers

Example: 12 = 2² × 3; 18 = 2 × 3² HCF = 2¹ × 3¹ = 6

Method 2: Euclidean Algorithm For HCF of a, b (a > b): divide a by b, replace a with b and b with remainder, repeat until remainder is 0.

LCM (Least Common Multiple):

Method 1: Prime factorisation LCM = product of prime factors with maximum powers

Example: 12 = 2² × 3; 18 = 2 × 3² LCM = 2² × 3² = 36

HCF and LCM Relationship: $$HCF(a,b) \times LCM(a,b) = a \times b$$

This is true for any two positive integers.

Finding Number of Factors:

If $N = 2^a \times 3^b \times 5^c …$ Then number of factors (divisors) of N = $(a+1)(b+1)(c+1)…$

Example: 12 = 2² × 3¹ Number of factors = (2+1)(1+1) = 3 × 2 = 6 Factors of 12 are: 1, 2, 3, 4, 6, 12

Sum of Factors: Sum of all factors of $N = (2^{a+1}-1)(3^{b+1}-1)(5^{c+1}-1)… / ((2-1)(3-1)(5-1)…)$

Remainder Theorems:

For polynomial P(x) divided by (x-a): Remainder = P(a)

Example: Find remainder when $x^2 + 3x + 2$ is divided by (x-1) P(1) = 1 + 3 + 2 = 6

For divisibility by specific numbers:

• Div by 2: Last digit even
• Div by 3: Sum of digits divisible by 3
• Pattern for remainders when dividing by 7, 11, 13: Use alternating group subtraction method for large numbers

⚡ SSC CGL-Specific Tip: For large numbers, find last 3 digits and test divisibility by 8. For 11, use alternating sum of digits. For questions involving “when divided by N, leave remainder R”, use modular arithmetic: the answer must be of the form N×k + R.

Common Student Mistakes:

• Forgetting 1 is neither prime nor composite
• Confusing HCF with LCM (choosing product of primes with max vs min powers)
• Not applying the correct divisibility rule

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Euclidean Algorithm (Extended):

To find HCF(a,b):

1. $a = bq + r$ where $0 \le r < b$
2. Replace $(a,b)$ with $(b,r)$
3. Repeat until $r = 0$

Example: HCF(72, 56) 72 = 56 × 1 + 16 56 = 16 × 3 + 8 16 = 8 × 2 + 0 HCF = 8

Chinese Remainder Theorem:

When a number leaves remainders $r_1, r_2, r_3$ when divided by $d_1, d_2, d_3$ (pairwise coprime divisors): Find N such that: $N \equiv r_1 \pmod{d_1}$ $N \equiv r_2 \pmod{d_2}$ $N \equiv r_3 \pmod{d_3}$

Unit Digit Cycles:

Power	Cycle for base ending in 2
2¹	2
2²	4
2³	8
2⁴	6
2⁵	2 (cycle repeats every 4)

For $7^n$: 7¹=7 (unit digit 7) 7²=49 (unit digit 9) 7³=343 (unit digit 3) 7⁴=2401 (unit digit 1) 7⁵=16807 (unit digit 7, repeats every 4)

Divisibility by 7, 11, 13:

For 7, 11, 13 together: Since $7 \times 11 \times 13 = 1001$: A number is divisible by 7, 11, and 13 if and only if it’s divisible by 1001. If 3528008 is divisible by 1001 (3528 - 008 = 3520… check calculation), then it’s divisible by all three.

Pattern Subtraction Method for 7, 11, 13: For 3524: 3 - 524 = -521 → 521 ÷ 7? No, not divisible.

Sum of First n Natural Numbers: $$S_n = \frac{n(n+1)}{2}$$

Sum of squares: $S_n^2 = \frac{n(n+1)(2n+1)}{6}$

Sum of cubes: $S_n^3 = \left[\frac{n(n+1)}{2}\right]^2$

Arithmetic Progression:

$n$th term: $a_n = a + (n-1)d$ Sum of n terms: $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n(a + l)}{2}$ (where l = last term)

Geometric Progression:

$n$th term: $a_n = ar^{n-1}$ Sum of n terms: $S_n = \frac{a(r^n - 1)}{r - 1}$ for r ≠ 1

Last Digit Problems:

To find last digit of $n^m$:

• If n ends in 0: last digit = 0
• If n ends in 1: last digit = 1
• If n ends in 2, 3, 7, 8: Find m mod 4; if m mod 4 = 0, use 6 for bases ending in 2, 3, 7, 8
• If n ends in 4 or 9: If m is odd, last digit = n mod 10; if m is even, last digit = (n mod 10)² mod 10
• If n ends in 5: last digit = 5 (if m > 0)

Sample SSC CGL Questions:

Q1: Find HCF of 120, 150, 200 Prime factors: 120 = 2³ × 3 × 5 150 = 2 × 3 × 5² 200 = 2³ × 5² HCF = 2¹ × 3⁰ × 5¹ = 10

Q2: Find LCM of 12, 18, 24 12 = 2² × 3 18 = 2 × 3² 24 = 2³ × 3 LCM = 2³ × 3² = 72

Q3: Find remainder when $5^{23}$ is divided by 6. Pattern: $5^1$ mod 6 = 5, $5^2$ mod 6 = 1, $5^3$ mod 6 = 5, $5^4$ mod 6 = 1 23 is odd → remainder = 5

Q4: Unit digit of $7^{423}$ 423 mod 4 = 3 (since 420 is divisible by 4) Unit digit cycle for 7: 7, 9, 3, 1 (repeats every 4) Position 3 → unit digit = 3

⚡ Advanced Tip: For questions like “find the largest number that always divides $n^3 - n$”, use the factorisation: $n^3 - n = n(n-1)(n+1)$. This is the product of three consecutive integers. Among any three consecutive integers, one is divisible by 3, and at least one is even, so product is always divisible by 6. Actually, for $n^3 - n$, it’s always divisible by 6, and in fact by 24 when n is odd (due to the product including consecutive even numbers).

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