Divisibility rules

Numbers are the atoms of mathematics — and before you can manipulate them in any exam, you must know exactly what kind of number you are dealing with, what rules govern it, and the sneaky shortcuts that shave minutes off a solution.

Definition — Natural numbers (N): positive counting numbers starting from 1: {1, 2, 3, 4, …}.

Definition — Whole numbers (W): natural numbers plus zero: {0, 1, 2, 3, …}.

Definition — Integers (Z): all positive and negative whole numbers and zero: {…, −3, −2, −1, 0, 1, 2, 3, …}.

Definition — Rational numbers (Q): numbers expressible as p/q where p, q are integers and q ≠ 0. Their decimal expansion either terminates (1/4 = 0.25) or repeats (1/3 = 0.333…).

Definition — Irrational numbers: real numbers that cannot be written as p/q. Their decimal expansions are non-terminating and non-repeating. Examples: √2, √3, π, e, √5.

Definition — Real numbers (R): the union of rational and irrational numbers — every point on the number line.

Definition — Complex numbers (C): numbers of the form a + ib where i = √(−1). They extend the real number line to a 2-D plane.

The Number Hierarchy

N ⊂ W ⊂ Z ⊂ Q ⊂ R ⊂ C

Each set extends the previous by including new kinds of numbers. This hierarchy is tested in MCQs like "√2 belongs to which set?" — answer: irrational ⊂ Real, but NOT rational or integer.

Key Algebraic Properties

Closure: A set is closed under an operation if performing that operation on two members always gives a member of the same set.

• Addition and multiplication are closed on N, W, Z, Q, R, C.
• Subtraction is not closed on N (3 − 5 = −2 ∉ N) but is on Z, Q, R, C.
• Division is not closed on Z (3 ÷ 2 = 1.5 ∉ Z) but is on Q{0}, R{0}, C{0}.

Associativity: (a + b) + c = a + (b + c) — true for addition and multiplication on all standard sets.

Commutativity: a + b = b + a and a × b = b × a. Subtraction and division are not commutative.

Distributivity: a × (b + c) = ab + ac. This is the bridge between addition and multiplication.

Identity elements: 0 is the additive identity (a + 0 = a). 1 is the multiplicative identity (a × 1 = a).

Inverse elements: −a is the additive inverse of a. 1/a is the multiplicative inverse of a (for a ≠ 0).

Divisibility Rules (Must-Memorise)

These rules let you test divisibility without performing division — critical in timed exams.

Divisor	Rule	Example
2	Last digit even	348 ✓
3	Sum of digits divisible by 3	123: 1+2+3=6 ✓
4	Last two digits divisible by 4	1732: 32÷4=8 ✓
5	Last digit 0 or 5	475 ✓
6	Divisible by both 2 AND 3	312: even, 3+1+2=6 ✓
7	Double the last digit, subtract from the rest; repeat if needed	343: 34 − 2×3 = 28 = 4×7 ✓
8	Last three digits divisible by 8	1,512: 512÷8=64 ✓
9	Digit sum divisible by 9	729: 7+2+9=18 ✓
10	Last digit 0	450 ✓
11	Alternating digit sum divisible by 11	1,364: (1+6)−(3+4)=0 ✓
12	Divisible by both 3 AND 4
25	Last two digits 00, 25, 50, or 75

Rule of 11 explained: Alternate + and − from the right: for 7,139: 9 − 3 + 1 − 7 = 0 → divisible by 11.

Prime Numbers

A prime is a natural number greater than 1 divisible only by 1 and itself.

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

2 is the only even prime. 1 is neither prime nor composite by definition.

Primality test for a number n: check divisibility by all primes up to √n. If none divide it, n is prime.

LCM and HCF

Definition — HCF (Highest Common Factor) / GCD: the largest number that divides all given numbers exactly.

Definition — LCM (Least Common Multiple): the smallest number divisible by all the given numbers.

Method 1 — Prime Factorisation:

• 18 = 2 × 3²; 24 = 2³ × 3.
• HCF: take the minimum power of each common prime → 2¹ × 3¹ = 6.
• LCM: take the maximum power of every prime that appears → 2³ × 3² = 72.

Method 2 — Euclidean Algorithm (for HCF):
gcd(a, b) = gcd(b, a mod b), repeating until remainder is 0.

Example: gcd(48, 18) → gcd(18, 12) → gcd(12, 6) → gcd(6, 0) = 6.

Golden property: For any two numbers a and b:

a × b = LCM(a, b) × HCF(a, b)

Example: 18 × 24 = 432 = 72 × 6 ✓

Co-prime numbers: HCF = 1. Example: 7 and 12 share no common factor except 1.

Remainder Theorem and Modular Arithmetic

Finding the remainder of large powers is a standard exam problem.

Cyclicity method: compute the first few powers and find the repeating pattern.

Example: Find 2^47 mod 7.

• 2¹ mod 7 = 2; 2² mod 7 = 4; 2³ mod 7 = 1; then repeats with cycle 3.
• 47 mod 3 = 2 → 2nd in cycle = 4.

Fermat's Little Theorem: If p is prime and gcd(a, p) = 1, then aᵖ⁻¹ ≡ 1 (mod p).

Example: 2⁶ ≡ 1 (mod 7) → 2^42 ≡ 1 (mod 7) → 2^47 = 2^42 × 2^5 → 1 × 32 mod 7 = 4. Consistent.

Wilson's Theorem: (p − 1)! ≡ −1 (mod p) for prime p.

Unit Digit (Last Digit) Patterns

Competitive exams love questions like "What is the unit digit of 7^100?"

The unit digit of powers follows a cycle:

Base	Cycle	Cycle length
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
0, 1, 5, 6	0, 1, 5, 6	1 (fixed)

Finding unit digit of 7^100:

• Cycle of 7 is (7, 9, 3, 1), length 4.
• 100 ÷ 4 = 25 remainder 0 → position 4 in cycle → unit digit = 1.

Rule: if exponent mod (cycle length) = 0, take the last element of the cycle.

Factorial and Trailing Zeros

Definition — Factorial (n!): n! = n × (n−1) × … × 2 × 1. Convention: 0! = 1.

Trailing zeros in n!: each zero requires one factor of 10 = 2 × 5. Since factors of 2 always exceed factors of 5 in n!, count the factors of 5 using Legendre's formula:

Number of trailing zeros = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + …

Example: trailing zeros in 100!
= 100/5 + 100/25 + 100/125 = 20 + 4 + 0 = 24

Example: trailing zeros in 200!
= 40 + 8 + 1 = 49

Surds and Indices

Definition — Surd: an irrational root expression that cannot be simplified to a rational number, e.g., √2, ∛5, √(2+√3).

Laws of Indices:

• aᵐ × aⁿ = aᵐ⁺ⁿ
• aᵐ / aⁿ = aᵐ⁻ⁿ
• (aᵐ)ⁿ = aᵐⁿ
• a⁰ = 1 (a ≠ 0)
• a⁻ⁿ = 1/aⁿ
• a^(1/n) = ⁿ√a
• a^(m/n) = (ⁿ√a)ᵐ

Rationalising the denominator: multiply numerator and denominator by the conjugate.

Example: 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/(3 − 1) = (√3 + 1)/2.

Worked Examples

Q1. Find LCM and HCF of 18 and 24.

• 18 = 2 × 3², 24 = 2³ × 3.
• HCF = 2 × 3 = 6; LCM = 2³ × 3² = 72.
• Check: 18 × 24 = 432 = 72 × 6 ✓.

Q2. Unit digit of 3^25?

• Cycle of 3: (3, 9, 7, 1), length 4. 25 mod 4 = 1 → 1st in cycle = 3.

Q3. Sum of all 2-digit numbers divisible by 7.

• Sequence: 14, 21, 28, …, 98. AP with first term a = 14, last term l = 98, d = 7.
• Number of terms n = (98 − 14)/7 + 1 = 84/7 + 1 = 12 + 1 = 13.
• Sum = n/2 × (a + l) = 13/2 × 112 = 728.

Q4. How many trailing zeros in 50!?

• 50/5 + 50/25 = 10 + 2 = 12 zeros.

Quick-revision notes for Divisibility rules — the must-know points for SSC CGL Tier-I/II.

• Divisibility rules let you check divisibility without full division — key for simplification and number problems.
• By 2: last digit even; by 5: last digit 0 or 5; by 10: last digit 0.
• By 3: sum of digits divisible by 3; by 9: sum of digits divisible by 9.
• By 4: last two digits divisible by 4; by 8: last three digits divisible by 8.
• By 11: (sum of digits in odd places) − (sum in even places) is 0 or a multiple of 11.
• By 6: divisible by both 2 and 3; combine rules for composite divisors (e.g. 12 = 3 and 4).

Cover the answer, recall, then check. 7 cards on the must-know Divisibility rules facts for SSC CGL.

Q1. State the divisibility rule for 3.
A1. A number is divisible by 3 if the sum of its digits is divisible by 3.

Q2. What is the divisibility rule for 8?
A2. If the number formed by its last three digits is divisible by 8.

Q3. State the divisibility rule for 11.
A3. If the difference between the sum of digits at odd places and even places is 0 or a multiple of 11.

Q4. How do you test divisibility by 4?
A4. Check if the last two digits form a number divisible by 4.

Q5. A number is divisible by 6 when?
A5. When it is divisible by both 2 and 3.

Q6. Is 918 divisible by 9? Why?
A6. Yes — digit sum 9+1+8 = 18, which is divisible by 9.

Q7. How do you test divisibility by 12?
A7. Check divisibility by both 3 and 4 (its coprime factors).

Key formulas

• 2: last digit even; 4: last two digits ÷4; 8: last three digits ÷8.
• 3: digit sum ÷3; 9: digit sum ÷9.
• 5: ends in 0 or 5; 10: ends in 0; 25: last two digits ÷25.
• 11: (sum of odd-place digits − sum of even-place digits) is 0 or ÷11.
• 6: divisible by 2 and 3; 12: by 3 and 4; composite rule = coprime factors.
• 7: subtract twice the last digit from the rest, repeat.

Usage: for a composite divisor, split into coprime factors and test each rule.

Worked Example

Problem: Find the least value of the digit x so that the 5-digit number 78x94 is divisible by 11.

Solution:
Divisibility rule for 11: (sum of digits at odd positions) − (sum at even positions) must be 0 or a multiple of 11. Number the positions from the right.
Digits from right: 4 (pos 1), 9 (pos 2), x (pos 3), 8 (pos 4), 7 (pos 5).
Odd positions (1, 3, 5): 4 + x + 7 = 11 + x
Even positions (2, 4): 9 + 8 = 17
Difference = (11 + x) − 17 = x − 6.
Set x − 6 = 0 → x = 6. (The next multiple would be x − 6 = 11 → x = 17, not a single digit.)

Answer: x = 6, giving the number 78694.