Number System

Free preview

Divisibility, factors, HCF, LCM, remainders.

Practice — 2976 questions

This is a free preview chapter. Unlock all of SSC CGL →

Divisibility rules

Rules for 2, 3, 4, 5, 6, 7, 8, 9, 11.

Number system — natural, rational, real, divisibility, LCM/HCF

Notes

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.

:::compare "Number Sets — Closure under Operations"

Set	+	−	×	÷
Natural (N)	✓	✗ (if result negative)	✓	✗
Whole (W)	✓	✗ (if result negative)	✓	✗
Integer (Z)	✓	✓	✓	✗
Rational (Q)	✓	✓	✓	✓ (except ÷0)
Real (R)	✓	✓	✓	✓ (except ÷0)
:::

:::keypoints Key points

N ⊂ W ⊂ Z ⊂ Q ⊂ R ⊂ C — every set contains the previous.
Rational numbers have terminating or repeating decimals; irrationals do not.
HCF × LCM = product of the two numbers (for two numbers only).
Unit digit cycles have length 4 for bases 2, 3, 7, 8 and length 2 for bases 4, 9.
Trailing zeros in n! = sum of ⌊n/5ᵏ⌋ for k = 1, 2, 3, …
Divisibility by 11: alternating digit sum; by 9: digit sum divisible by 9.
Co-prime numbers have HCF = 1 — useful for fraction simplification.
Fermat's Little Theorem speeds up large-power remainder calculations.
:::

:::memory
"No Work In Quiet Rest, Completely" for the number set hierarchy: Natural → Whole → Integer → Quotient (Rational) → Real → Complex.
For unit digits of powers: "2-3-7-8 cycle four, 4-9 cycle two, 0-1-5-6 stuck like glue."
:::

:::recap

Number hierarchy N ⊂ W ⊂ Z ⊂ Q ⊂ R ⊂ C; rationals have repeating/terminating decimals.
Divisibility shortcuts save time: rule of 9 (digit sum), rule of 11 (alternating sum), rule of 8 (last three digits).
HCF by prime factorisation (min powers) and LCM (max powers); their product = product of the two numbers.
Trailing zeros in n! = count factors of 5 using Legendre's formula.
Unit digit follows a cycle of length 4 for 2, 3, 7, 8; length 2 for 4, 9; fixed for 0, 1, 5, 6.
Rationalise surds by multiplying by the conjugate; use laws of indices to simplify.
:::

Types of Numbers

Classification of Numbers

Notes

Every number you encounter in mathematics — from the price of a bus ticket to the digits of π — belongs to one or more carefully defined families, and knowing these families is the foundation of every quantitative question in SSC CGL and similar competitive exams.

Definition: A number system is a structured classification of numbers into sets based on their properties, such as whether they can be expressed as a ratio, whether they have a repeating decimal expansion, and how they behave under arithmetic operations.

Natural Numbers (N)

Definition: The set of natural numbers is N = {1, 2, 3, 4, 5, …}. These are the counting numbers — the ones a child first learns to count objects with.

Key properties:

Smallest natural number: 1
No upper bound — they continue infinitely
Closed under addition and multiplication (adding or multiplying two natural numbers always gives a natural number)
NOT closed under subtraction (3 − 5 = −2, which is not in N)

Why it matters: Every counting operation — ranking candidates in an exam, counting factories in a district — involves natural numbers.

Whole Numbers (W)

Definition: W = {0, 1, 2, 3, 4, …} — natural numbers plus zero.

Smallest whole number: 0
Zero is the additive identity: a + 0 = a for any whole number a
Every natural number is also a whole number, but 0 is not a natural number

Common misconception: Many students treat "natural numbers" and "whole numbers" as synonyms. They are not — the difference is the presence of zero.

Integers (Z)

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

Integers are closed under addition, subtraction, and multiplication
Division of two integers need not be an integer (7 ÷ 2 = 3.5)
The letter Z comes from the German word Zahlen (numbers)

Real-world example: Temperature in Shimla in January can be −5°C, −3°C, or 2°C — all integers.

Rational Numbers (Q)

Definition: A rational number is any number that can be written in the form p/q, where p and q are integers and q ≠ 0.

Examples: 3/4, −7/2, 0.5 (= 1/2), 0.333… (= 1/3), 6 (= 6/1)

Key insight — decimal behaviour:

A rational number always has either a terminating decimal (e.g., 1/4 = 0.25) or a recurring (repeating) decimal (e.g., 1/3 = 0.333…)
If you see a decimal that terminates or repeats, it is rational

Why it matters: Fractions appear in ratio, proportion, percentage, and speed-distance problems — all high-frequency SSC topics.

Irrational Numbers

Definition: An irrational number cannot be expressed as p/q for any integers p and q. Its decimal expansion is non-terminating and non-repeating.

Common irrational numbers to memorise:

√2 ≈ 1.41421…
√3 ≈ 1.73205…
√5 ≈ 2.23606…
π ≈ 3.14159…
e ≈ 2.71828…

Common misconception: Students often think 22/7 = π. It does NOT — 22/7 is a rational approximation; π itself is irrational.

Another trap: √4 = 2, which IS rational. Not every square root is irrational — only the square root of a non-perfect-square is irrational.

Real Numbers (R)

Definition: The set of real numbers is the union of all rational and irrational numbers. Every point on the number line corresponds to exactly one real number.

:::compare Subsets of Real Numbers

Set	Notation	Contains	Example
Natural	N	{1, 2, 3, …}	7
Whole	W	{0, 1, 2, 3, …}	0
Integer	Z	{…−2,−1, 0, 1, 2…}	−5
Rational	Q	p/q, q≠0	3/4
Irrational	—	Non-repeating decimals	√2
Real	R	Q ∪ Irrational	All of the above
:::

The hierarchy: N ⊂ W ⊂ Z ⊂ Q ⊂ R

Prime Numbers

Definition: A prime number has exactly two distinct positive factors: 1 and itself.

Important facts:

2 is the only even prime number
1 is NOT prime (it has only one factor: itself)
The number of primes below 100 is 25

List of all 25 primes below 100:
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

Why it matters: Prime factorisation is the engine behind LCM, HCF, number of divisors, and many SSC number theory problems.

Composite Numbers

Definition: A composite number has more than two factors — i.e., it can be divided by at least one number other than 1 and itself.

Examples: 4 (factors: 1, 2, 4), 12 (factors: 1, 2, 3, 4, 6, 12)

Key fact: Every composite number can be expressed as a unique product of primes (Fundamental Theorem of Arithmetic).

Example: 360 = 2³ × 3² × 5

Special case: 1 is neither prime nor composite. This is one of the most frequently tested tricky facts in SSC CGL.

Co-prime Numbers

Definition: Two numbers are co-prime (or relatively prime) if their HCF = 1 — they share no common factor other than 1.

Examples:

8 and 15: factors of 8 = {1, 2, 4, 8}; factors of 15 = {1, 3, 5, 15}; HCF = 1 → co-prime ✓
4 and 9: HCF = 1 → co-prime ✓
6 and 9: HCF = 3 → NOT co-prime ✗

Key insight: Co-prime numbers need not be prime themselves — 8 and 15 are co-prime even though neither is prime.

Why it matters: Co-primality appears in LCM/HCF questions and in fraction simplification problems.

Even and Odd Numbers

Even numbers: Divisible by 2. Last digit is 0, 2, 4, 6, or 8.
Odd numbers: Not divisible by 2. Last digit is 1, 3, 5, 7, or 9.

Arithmetic rules (heavily tested in SSC):

Even + Even = Even
Odd + Odd = Even
Even + Odd = Odd
Even × Even = Even
Odd × Odd = Odd
Even × Odd = Even

Perfect Numbers

Definition: A perfect number equals the sum of all its proper divisors (all divisors except itself).

Example: 6 → proper divisors: 1, 2, 3 → sum = 6 ✓
Example: 28 → proper divisors: 1, 2, 4, 7, 14 → sum = 28 ✓

The next perfect number after 28 is 496. For SSC purposes, just remember 6 and 28.

Worked Example — Classifying a Number

Question: Classify the number 0.171717… and determine which sets it belongs to.

Solution:

Step 1: The decimal 0.1717… has a repeating block "17". A non-terminating but repeating decimal is always rational.

Step 2: Express as fraction: Let x = 0.1717…
Then 100x = 17.1717…
100x − x = 17 → 99x = 17 → x = 17/99

Step 3: 17/99 is in the form p/q (p=17, q=99, q≠0).

Conclusion: 0.1717… is a rational number, and therefore also a real number. It is not a natural number, whole number, or integer.

:::keypoints Key points

Natural numbers start at 1; whole numbers start at 0; integers include negatives
Every natural ⊂ whole ⊂ integer ⊂ rational ⊂ real
Rational decimals terminate or repeat; irrational decimals never repeat
2 is the only even prime number
1 is neither prime nor composite — this is a favourite SSC trap
Co-prime means HCF = 1, not that either number is prime
There are exactly 25 prime numbers less than 100
Perfect numbers: 6, 28 (sum of proper divisors equals the number)
:::

:::memory
"Never Will I Quit Reaching" → Natural, Whole, Integer, Rational, Real — each set contains the previous one. For primes: "2 is the lone ranger" — the only even prime.
:::

:::recap

Numbers split into Natural → Whole → Integer → Rational → Real (each is a subset of the next)
Rational numbers have terminating or recurring decimals; irrational numbers have neither
Prime: exactly 2 factors; Composite: more than 2; 1: neither
Co-prime pairs have HCF = 1 (e.g., 8 and 15)
Even/odd arithmetic rules are directly tested in SSC CGL
25 primes exist below 100; memorise the full list
:::

Key Properties of Special Numbers

Notes

Certain numbers have such special mathematical properties that they appear as traps and shortcuts in competitive exams again and again — mastering them means converting hard questions into one-step recalls.

Prime Numbers up to 100

Every serious SSC CGL aspirant must have the complete list of 25 primes below 100 memorised:

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

Why memorise the full list? SSC CGL regularly asks "which of the following is prime?" with options like 91, 97, 87, 93 — if you know the list, the answer is instant.

Grouping trick for memory:

Teens primes: 11, 13, 17, 19 (all four — skip 15)
Twenties primes: only 23 and 29
Thirties primes: 31, 37 (only two)
Forties primes: 41, 43, 47 (three)
Fifties primes: 53, 59 (only two — skip 51=3×17, 55=5×11, 57=3×19)
Sixties primes: 61, 67 (only two — 63=9×7, 65=5×13)
Seventies primes: 71, 73, 79 (three)
Eighties primes: 83, 89 (only two — 81=3⁴, 87=3×29)
Ninety primes: 97 only (91=7×13 is a classic trap!)

How to Test if a Number is Prime

Definition: To check if n is prime, test divisibility by all prime numbers up to √n.

Why only up to √n? If n has a factor larger than √n, then its corresponding paired factor must be smaller than √n — so you would have found it already.

Worked illustration:

Test n = 221. √221 ≈ 14.8. Check primes: 2, 3, 5, 7, 11, 13.
- 221 ÷ 13 = 17. So 221 = 13 × 17. Composite.
Test n = 233. √233 ≈ 15.3. Check 2, 3, 5, 7, 11, 13: none divide 233. Prime.

Perfect Numbers

Definition: A perfect number is a positive integer that equals the sum of all its proper divisors (all positive divisors excluding the number itself).

The first two perfect numbers (and the only ones relevant for SSC CGL):

Number	Proper Divisors	Sum
6	1, 2, 3	6 ✓
28	1, 2, 4, 7, 14	28 ✓

The next perfect number is 496, then 8128 — far beyond competitive exam scope.

Why it matters: "Which of the following is a perfect number?" is a direct recall question.

Common misconception: Students sometimes confuse "perfect number" with "perfect square." A perfect square is n² for integer n (4, 9, 16, 25…). These are entirely different concepts.

Twin Primes

Definition: Twin primes are pairs of prime numbers that differ by exactly 2.

Known twin prime pairs (up to 100):

(3, 5)
(5, 7)
(11, 13)
(17, 19)
(29, 31)
(41, 43)
(59, 61)
(71, 73)

Note: (2, 3) differ by 1, not 2 — not twin primes.

Twin Prime Conjecture: It is believed (but not proven) that there are infinitely many twin prime pairs. This is one of the great unsolved problems in mathematics.

Even and Odd Number Properties

These arithmetic rules are tested both directly and embedded in larger problems:

:::compare Even-Odd Arithmetic Rules

Operation	Result	Example
Even + Even	Even	4 + 6 = 10
Odd + Odd	Even	3 + 7 = 10
Even + Odd	Odd	4 + 3 = 7
Even × Even	Even	4 × 6 = 24
Odd × Odd	Odd	3 × 7 = 21
Even × Odd	Even	4 × 3 = 12
Even − Even	Even	8 − 4 = 4
Odd − Odd	Even	7 − 3 = 4
:::

Key insight for SSC: The product of ANY set of numbers is even if at least one of them is even. The product is odd only if all numbers in the product are odd.

Real-world example: If you buy 3 items priced at ₹5, ₹7, and ₹9 each (all odd), total = ₹21 (odd). Add one item at ₹4 (even) and the product of prices is now even.

Properties of Powers (Even/Odd Exponents)

For SSC number theory and remainder problems:

Even number raised to any positive power → Even
Odd number raised to any positive power → Odd
The last digit of a power follows a cycle (tested in cyclicity problems)

Last digit cycles for common bases:

2: cycle 2, 4, 8, 6 (period 4)
3: cycle 3, 9, 7, 1 (period 4)
7: cycle 7, 9, 3, 1 (period 4)
8: cycle 8, 4, 2, 6 (period 4)
1, 5, 6: last digit never changes (always 1, 5, 6 respectively)

Composite Number Properties

Every composite number n can be written as a product of primes in exactly one way (up to order) — this is the Fundamental Theorem of Arithmetic.

Number of divisors formula: If n = p₁^a × p₂^b × p₃^c …, then number of divisors = (a+1)(b+1)(c+1)…

Example: 360 = 2³ × 3² × 5¹
Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24

Worked Example — Applying Multiple Properties

Question: The product of two prime numbers is 77. What is their sum?

Solution:

Step 1: Express 77 as a product. 77 = 7 × 11.

Step 2: Check: 7 is prime (not divisible by 2, 3, 5; √7 < 3). 11 is prime.

Step 3: Sum = 7 + 11 = 18.

Question: How many prime numbers are there between 50 and 70?

Solution:

Step 1: List candidates: 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.

Step 2: Eliminate composites: 51=3×17, 52=2×26, 54=2×27, 55=5×11, 56=2×28, 57=3×19, 58=2×29, 60=2×30, 62=2×31, 63=7×9, 64=2⁶, 65=5×13, 66=2×33, 68=2×34, 69=3×23.

Step 3: Remaining primes: 53, 59, 61, 67 → 4 primes.

:::keypoints Key points

There are 25 primes below 100; 91=7×13 is a notorious non-prime trap
To test primality of n, check divisibility by primes up to √n only
Perfect numbers: 6 and 28 (proper divisors sum to the number)
Twin primes differ by exactly 2; pairs up to 100: (3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61),(71,73)
Product of numbers is odd ONLY IF every factor is odd
Sum of two odd numbers is always even
Number of divisors of n = p^a × q^b is (a+1)(b+1)
:::

:::memory
"Perfect: Six and Twenty-Eight" — 6 = 1+2+3; 28 = 1+2+4+7+14. For twin primes: think of prime siblings sitting next to each other on the number line with one seat gap.
:::

:::recap

Full prime list below 100 has exactly 25 numbers; memorise by decade groups
Primality test: divide by primes ≤ √n; if none divide, it is prime
Perfect number = sum of proper divisors; 6 and 28 are the exam-relevant ones
Twin primes are prime pairs with difference 2
Product is even if any factor is even; product is odd only if all factors are odd
Number of divisors uses the prime factorisation exponent formula (a+1)(b+1)…
:::

Solved Example: Identifying Number Types

Worked example

The fastest route to a correct answer in SSC CGL is not just knowing the definition of a prime number — it's being able to identify one (or rule it out) in under 30 seconds without a calculator.

The Core Technique: Square Root Boundary Test

Definition: A number n is prime if and only if it has no prime factor less than or equal to √n.

This works because:

If n = a × b and both a > √n and b > √n, then a × b > n — contradiction.
So at least one factor must be ≤ √n.
Therefore, if no prime ≤ √n divides n, then n has no factor at all (other than 1 and n), so it is prime.

Practical rule:

For numbers up to 100: check divisibility by 2, 3, 5, 7 (since √100 = 10, and 7 is the largest prime below 10)
For numbers up to 200: also check 11 and 13 (since √200 ≈ 14.1)
For numbers up to 400: also check 17 and 19 (since √400 = 20)

Divisibility Quick-Checks

Before doing division, use divisibility rules to eliminate candidates instantly:

Divisor	Rule	Example
2	Last digit even	84: last digit 4 → divisible
3	Sum of digits divisible by 3	93: 9+3=12 → divisible
5	Last digit 0 or 5	95: last digit 5 → divisible
7	No simple rule; just divide
11	(Sum of odd-position digits) − (Sum of even-position digits) = 0 or 11	121: (1+1)−2=0 → divisible
13	No simple rule; just divide

Solved Example 1 (from the lesson stub)

Question: Which of the following is a prime number — 91, 97, 87, 93?

Solution:

Step 1: Examine 91. √91 ≈ 9.5 → check primes 2, 3, 5, 7.

Is 91 even? No.
Digit sum = 9+1 = 10 → not divisible by 3.
Last digit not 0 or 5 → not divisible by 5.
91 ÷ 7 = 13. Yes! So 91 = 7 × 13. Composite.

Step 2: Examine 87. √87 ≈ 9.3 → check 2, 3, 5, 7.

Digit sum = 8+7 = 15 → divisible by 3. 87 = 3 × 29. Composite.

Step 3: Examine 93. √93 ≈ 9.6 → check 2, 3, 5, 7.

Digit sum = 9+3 = 12 → divisible by 3. 93 = 3 × 31. Composite.

Step 4: Examine 97. √97 ≈ 9.8 → check 2, 3, 5, 7.

Not even. Digit sum = 9+7 = 16 → not divisible by 3. Last digit ≠ 0,5. 97 ÷ 7 = 13.857… Not divisible.
No prime ≤ √97 divides 97. 97 is prime.

Conclusion: 97 is the prime.

Why 91 is the classic trap: 91 looks prime because it is not divisible by 2, 3, or 5. But 7 × 13 = 91 — always check 7 for two-digit numbers!

Solved Example 2: Identifying the Type of a Given Number

Question: Classify: 1, 2, 4, 15, 17, 28, 36.

Solution:

Number	Factors	Classification
1	1 only	Neither prime nor composite
2	1, 2	Prime (only even prime)
4	1, 2, 4	Composite
15	1, 3, 5, 15	Composite
17	1, 17	Prime
28	1, 2, 4, 7, 14, 28	Composite; also a perfect number
36	1,2,3,4,6,9,12,18,36	Composite; perfect square (6²)

Solved Example 3: How Many Primes Between Two Numbers?

Question: How many prime numbers lie between 80 and 100?

Solution:

Step 1: Candidates: 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99.

Step 2: Eliminate composites quickly —

Even numbers: 82, 84, 86, 88, 90, 92, 94, 96, 98 → all composite
Divisible by 5: 85 (=5×17), 90, 95 (=5×19) → composite
Divisible by 3: 81 (=3⁴), 87 (=3×29), 93 (=3×31), 99 (=9×11) → composite
91 = 7 × 13 → composite

Step 3: Remaining: 83 and 89 and 97.

83: √83 ≈ 9.1. Not divisible by 2, 3, 5, 7. → Prime
89: √89 ≈ 9.4. Not divisible by 2, 3, 5, 7. → Prime
97: already confirmed prime above.

Conclusion: 3 primes between 80 and 100 (83, 89, 97).

Common SSC Traps

Trap 1 — 1 is prime: FALSE. 1 is neither prime nor composite.

Trap 2 — 91 is prime: FALSE. 91 = 7 × 13.

Trap 3 — 2 is not prime (it's even): FALSE. 2 is the only even prime.

Trap 4 — All odd numbers are prime: FALSE. 9 = 3×3, 15 = 3×5, 21 = 3×7 are all odd composites.

Trap 5 — √n must be an integer to be relevant: FALSE. Even if √n is not an integer (e.g., √91 ≈ 9.5), you still check all primes up to the floor of √n.

Worked Shortcut: Composite Identification by Digit Sum

Question: Without dividing, determine whether 561 is prime.

Solution:

Step 1: Digit sum = 5+6+1 = 12. Divisible by 3 → 561 is divisible by 3. Composite. (561 = 3 × 187)

This digit-sum trick eliminates most large numbers before you even attempt division.

:::keypoints Key points

To test primality of n, check only primes ≤ √n — once you pass √n, stop
For numbers 1–100: checking 2, 3, 5, 7 is sufficient (√100 = 10)
91 = 7 × 13 is the most common prime-impersonator in SSC exams
1 is neither prime nor composite — a guaranteed trick question
Use divisibility rules (digit sum for 3; last digit for 2, 5) before attempting division
Between 80 and 100 there are exactly 3 primes: 83, 89, 97
Perfect numbers (6, 28) and twin primes are separate special categories
:::

:::memory
"Seven Traps Ninety-One" — whenever you see 91, immediately check ÷7. For the test: "Check up to the Square Root, not a step more."
:::

:::recap

Primality test: find √n, then test divisibility by each prime up to that bound
Apply divisibility rules first (2, 3, 5) to quickly eliminate most candidates
91 = 7×13, 87 = 3×29, 93 = 3×31 — all favourite composite traps in MCQ options
97 is the largest prime below 100
1 is not prime; 2 is prime and is the only even prime
Between any two consecutive perfect squares, count primes carefully using the method above
:::

HCF and LCM

Prime factorization, division method, product = HCF × LCM.

HCF, LCM and divisibility rules — the foundations

Notes

Number theory might look like pure abstraction, but HCF and LCM are the silent engines behind timetable synchronisation, tile-laying puzzles, cutting material without waste, and nearly every quantitative-aptitude section in SSC CGL, RRB NTPC, and IBPS exams.

Foundational Concepts

Definition: A factor (or divisor) of a number n is any integer that divides n exactly (with no remainder). Every number has 1 and itself as factors.

Definition: A multiple of n is any integer that is exactly divisible by n: multiples of 6 are 6, 12, 18, 24, ...

Definition: A prime number is a natural number greater than 1 with exactly two factors: 1 and itself. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 ...

Definition: HCF (Highest Common Factor) — also called GCD (Greatest Common Divisor) — is the largest number that divides each of the given numbers exactly.

Definition: LCM (Least Common Multiple) is the smallest positive number that is divisible by each of the given numbers.

Divisibility Rules — The Speed Engine

Knowing these rules lets you instantly check whether large numbers are divisible without actual division.

:::compare Divisibility rules

Divisor	Rule	Example
2	Last digit is even (0,2,4,6,8)	4,726 — last digit 6 ✓
3	Sum of all digits divisible by 3	5,271: 5+2+7+1=15 ✓
4	Last 2 digits divisible by 4	2,348: 48÷4=12 ✓
5	Last digit is 0 or 5	3,445 ✓
6	Divisible by BOTH 2 and 3	2,412: even ✓, digit sum 9 ✓
7	Double last digit, subtract from rest; if result is 0 or divisible by 7	343: double 3=6; 34−6=28; 28÷7=4 ✓
8	Last 3 digits divisible by 8	4,096: 096÷8=12 ✓
9	Sum of all digits divisible by 9	7,281: 7+2+8+1=18 ✓
10	Last digit is 0	5,730 ✓
11	Alternating sum (odd positions − even positions) divisible by 11	3,729: (3+2)−(7+9)=5−16=−11 ✓
12	Divisible by both 3 and 4	2,604: digit sum 12 ✓, last two 04 ✓
25	Last 2 digits are 00, 25, 50, or 75	4,375: last two 75 ✓
:::

Rule of 7 — worked step-by-step: Is 161 divisible by 7?

Last digit = 1; double it = 2.
Remaining number = 16.
16 − 2 = 14.
14 ÷ 7 = 2. Yes, divisible.

Rule of 11 — worked step-by-step: Is 91,861 divisible by 11?

Odd positions (from left): 9, 8, 1 → sum = 18.
Even positions: 1, 6 → sum = 7.
Alternating difference: 18 − 7 = 11. Divisible by 11. ✓

Why these rules matter: In multi-part problems, checking divisibility saves you from calculating long divisions. Especially useful for factor/LCM problems and "how many numbers between X and Y are divisible by Z" questions.

Finding HCF — Two Methods

Method 1: Prime Factorisation

Write each number as a product of prime factors in exponential form. HCF = product of minimum powers of all common primes.

Question: Find HCF(72, 120, 180).
Solution:
Step 1: 72 = 2³ × 3²; 120 = 2³ × 3 × 5; 180 = 2² × 3² × 5.
Step 2: Common primes: 2 and 3.
Step 3: Min power of 2 = 2²; min power of 3 = 3¹.
Conclusion: HCF = 2² × 3 = 4 × 3 = 12.

Method 2: Euclidean Algorithm (fastest for large numbers)

Based on the property: HCF(a, b) = HCF(b, a mod b). Repeat until remainder = 0.

Question: Find HCF(252, 105).
Solution:
Step 1: 252 = 2 × 105 + 42 → HCF(252, 105) = HCF(105, 42).
Step 2: 105 = 2 × 42 + 21 → HCF(105, 42) = HCF(42, 21).
Step 3: 42 = 2 × 21 + 0 → HCF(42, 21) = 21.
Conclusion: HCF(252, 105) = 21.

Finding LCM — Two Methods

Method 1: Prime Factorisation

LCM = product of maximum powers of all primes appearing in any of the numbers.

Question: Find LCM(72, 120, 180).
Solution:
Step 1: 72 = 2³ × 3²; 120 = 2³ × 3 × 5; 180 = 2² × 3² × 5.
Step 2: Max power of 2 = 2³; max power of 3 = 3²; max power of 5 = 5¹.
Conclusion: LCM = 8 × 9 × 5 = 360.

Method 2: Using HCF

For two numbers only: LCM(a, b) = (a × b) / HCF(a, b).

Example: LCM(48, 18) = (48 × 18) / HCF(48, 18) = 864 / 6 = 144.

Important warning: The formula LCM × HCF = a × b applies ONLY for two numbers. For three or more numbers, this shortcut does NOT hold.

The Golden Relationship

HCF(a, b) × LCM(a, b) = a × b (for two numbers)

This identity is used to find one of HCF/LCM when the other and the product are known.

HCF and LCM of Fractions

	Formula
HCF of fractions	HCF(numerators) / LCM(denominators)
LCM of fractions	LCM(numerators) / HCF(denominators)

Example: HCF(3/4, 9/16, 6/8):

HCF(3, 9, 6) = 3; LCM(4, 16, 8) = 16.
HCF of fractions = 3/16.

Key Exam Problem Types

Type 1: Remainder Problems — Same Remainder

"Find the greatest number that divides 250, 435, and 630 leaving the same remainder."

Method: HCF of differences: |435 − 250| = 185; |630 − 435| = 195; |630 − 250| = 380.
HCF(185, 195, 380).
185 = 5 × 37; 195 = 5 × 39 = 5 × 3 × 13; 380 = 4 × 5 × 19.
HCF = 5.
Answer: 5.

Type 2: Smallest Number with Given Remainder — Same Divisors

"Find the smallest number that when divided by 6, 8, and 15 leaves a remainder of 2."

Method: LCM(6, 8, 15) + 2.
LCM(6, 8, 15): 6=2×3; 8=2³; 15=3×5. LCM = 2³ × 3 × 5 = 120.
Answer: 120 + 2 = 122.

Type 3: Trailing Zeros in n!

The number of trailing zeros in n! equals the number of times 10 is a factor of n! = min(power of 2, power of 5). Since powers of 2 always exceed powers of 5:

Trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

Question: How many trailing zeros in 100!?
Solution: ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24.

Question: How many trailing zeros in 50!?
Solution: ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2 = 12.

Type 4: Number of Factors

If n = p₁^a × p₂^b × p₃^c × ..., then:

Number of factors = (a+1)(b+1)(c+1)...
Sum of factors = [(p₁^(a+1) − 1)/(p₁ − 1)] × [(p₂^(b+1) − 1)/(p₂ − 1)] × ...

Question: Find the number of factors of 360.
Solution:
360 = 2³ × 3² × 5¹.
Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

Question: Find the sum of factors of 72.
Solution:
72 = 2³ × 3².
Sum = [(2⁴−1)/(2−1)] × [(3³−1)/(3−1)] = [15/1] × [26/2] = 15 × 13 = 195.

Type 5: Bells / Lights / Signals Blinking Together

"Three bells ring at intervals of 6, 8, and 12 minutes. If they ring together at 7:00 AM, when will they next ring together?"

Method: LCM(6, 8, 12) = 24 minutes.
Answer: Next ring together at 7:24 AM.

Real-world example: Traffic light synchronisation in cities uses LCM logic — signal timings are chosen so that multiple lights re-synchronise at a predictable time, minimising bunching.

Type 6: Largest Tile / Plank / Ruler

"Find the largest square tile that can exactly tile a floor of 216 cm × 180 cm."

Method: HCF(216, 180) = ?
216 = 2³ × 3³; 180 = 2² × 3² × 5.
HCF = 2² × 3² = 4 × 9 = 36 cm.
The tile is 36 cm × 36 cm.

Properties and Useful Identities

HCF(a, b) always divides LCM(a, b).
HCF(a, b) divides every linear combination ma + nb (Bezout's identity).
If HCF(a, b) = 1, a and b are coprime (no common factor other than 1). LCM = a × b.
HCF(a, a+1) = 1 always (consecutive integers are always coprime).
HCF(a², b²) = [HCF(a, b)]²; LCM(a², b²) = [LCM(a, b)]².

Common Misconceptions

Common misconception: "HCF is always smaller than both numbers." True — HCF ≤ min(a, b). But students sometimes compute LCM and report it as HCF.

Common misconception: "LCM × HCF = product holds for any number of terms." FALSE — it holds only for exactly two numbers. For 3 numbers a, b, c: LCM × HCF ≠ a × b × c in general.

Common misconception: "Every common factor of two numbers is their HCF." Only the LARGEST common factor is the HCF.

:::keypoints Key points

Divisibility rules for 2–12 and 25 enable rapid mental arithmetic without full division.
HCF (prime factorisation): take minimum powers of common primes; Euclidean algorithm is faster for large numbers.
LCM (prime factorisation): take maximum powers of all primes in any of the numbers.
LCM × HCF = a × b — valid for exactly TWO numbers only.
"Smallest number leaving remainder r when divided by a, b, c" = LCM(a,b,c) + r.
"Largest number dividing a, b, c with same remainder" = HCF of pairwise differences.
Trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...
Number of factors of n = p^a × q^b × ... is (a+1)(b+1)...
:::

:::memory
"HCF = Minimum, LCM = Maximum" — in prime factorisation, HCF picks the minimum power of each prime, LCM picks the maximum. The product HCF × LCM = a × b works only for two numbers — think of a see-saw balanced between two weights.
:::

:::recap

HCF is the largest number that divides all given numbers; LCM is the smallest number divisible by all given numbers.
Prime factorisation method: HCF uses minimum powers of common primes; LCM uses maximum powers of all primes.
Euclidean algorithm: HCF(a, b) = HCF(b, a mod b) — fastest method for large numbers.
Bells/tiles/synchronisation problems → LCM; cutting without waste/common ruler problems → HCF.
Trailing zeros in n! counts factors of 5 using Legendre's formula.
LCM × HCF = a × b is valid only for two numbers — a common exam trap extends this incorrectly to three.
:::

Number system — natural, rational, real, divisibility, LCM/HCF

Notes

Every number you will ever meet in mathematics belongs to a nested family of sets, and knowing exactly which family it belongs to tells you what operations are safe to perform and how to solve the problems built around it.

Definition: The number system is the organised hierarchy of sets — Natural, Whole, Integer, Rational, Irrational, Real, Complex — where each larger set contains all the smaller ones plus new elements.

The hierarchy of number sets

:::compare Number Set Hierarchy

Set	Symbol	Members	Example
Natural numbers	N	1, 2, 3, …	Counting things
Whole numbers	W	0, 1, 2, 3, …	N + zero
Integers	Z	…, −2, −1, 0, 1, 2, …	W + negatives
Rational numbers	Q	p/q, q ≠ 0	3/4, −5, 0.25, 0.333…
Irrational numbers	—	cannot be written as p/q	√2, π, e
Real numbers	R	rational ∪ irrational	all points on the number line
Complex numbers	C	a + ib, i = √(−1)	3 + 4i
:::

A rational number's decimal expansion either terminates (1/4 = 0.25) or repeats (1/3 = 0.333…). An irrational number does neither — its decimal goes on forever without any repeating block.

Special subsets worth memorising

Prime numbers: exactly two distinct divisors (1 and itself): 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
Composite numbers: more than two divisors (4, 6, 8, 9, …).
1: neither prime nor composite.
2: the only even prime.
Co-prime (relatively prime): two numbers whose HCF = 1. Example: 7 and 12 share no factor other than 1.

Properties of operations

Closure — an operation is closed on a set if the result is always in the same set:

N is closed under + and × but not under − (3 − 5 = −2 ∉ N) or ÷ (3 ÷ 7 ∉ N).
Z is closed under +, −, × but not ÷ (7 ÷ 3 ∉ Z).
Q is closed under +, −, ×, and ÷ (excluding division by 0).

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

Associativity: (a + b) + c = a + (b + c). True for + and ×; not for − or ÷.

Distributivity: a × (b + c) = ab + ac.

Identities: 0 is the additive identity; 1 is the multiplicative identity.

Inverses: −a is the additive inverse of a; 1/a is the multiplicative inverse (a ≠ 0).

Divisibility rules

:::compare Divisibility Tests

Divisor	Rule	Example
2	Last digit even	346 ✓
3	Digit sum divisible by 3	123 → 1+2+3=6 ✓
4	Last two digits divisible by 4	312 → 12÷4=3 ✓
5	Last digit 0 or 5	235 ✓
6	Divisible by both 2 and 3	126 ✓
7	Double the last digit, subtract from rest; repeat	133 → 13−6=7 ✓
8	Last three digits divisible by 8	1,024 → 024÷8=3 ✓
9	Digit sum divisible by 9	729 → 7+2+9=18 ✓
10	Last digit is 0	340 ✓
11	Alternating digit sum divisible by 11	1,001 → 1−0+0−1=0 ✓
:::

LCM and HCF

LCM (Least Common Multiple): the smallest positive integer divisible by both numbers. Used in problems involving cycles, scheduling, or finding when two periodic events coincide.

HCF / GCD (Highest Common Factor): the largest integer that divides both numbers without a remainder. Used in problems involving splitting into equal groups or simplifying fractions.

Golden property: LCM(a, b) × HCF(a, b) = a × b

Method 1 — Prime factorisation (preferred for exam)

Example: HCF and LCM of 18 and 24.

18 = 2¹ × 3²
24 = 2³ × 3¹
HCF: take minimum power of each prime → 2¹ × 3¹ = 6
LCM: take maximum power of each prime → 2³ × 3² = 72
Check: 6 × 72 = 432 = 18 × 24 ✓

Method 2 — Euclidean algorithm (fast for two large numbers)

gcd(a, b) = gcd(b, a mod b), repeating until the remainder is 0.
Example: gcd(48, 18) → gcd(18, 12) → gcd(12, 6) → gcd(6, 0) = 6.

Remainder and modular arithmetic

For cyclic remainder problems, find the cycle length and use modular arithmetic.

Example: Find 2^10 mod 7.
Powers of 2 mod 7: 2, 4, 1, 2, 4, 1, … (cycle of length 3).
10 mod 3 = 1. So 2^10 mod 7 = 2¹ mod 7 = 2.

Fermat's little theorem: If p is prime and gcd(a, p) = 1, then a^(p−1) ≡ 1 (mod p). Useful for large exponents with prime moduli.

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

Unit-digit (last digit) patterns

The last digit of powers cycles with short periods:

:::compare Unit-digit cycles

Base	Cycle	Cycle length
0, 1, 5, 6	always same digit	1
4, 9	4,6,4,6 / 9,1,9,1	2
2	2,4,8,6	4
3	3,9,7,1	4
7	7,9,3,1	4
8	8,4,2,6	4
:::

Question: Unit digit of 7^100?
Cycle for 7: 7, 9, 3, 1 (length 4). 100 mod 4 = 0 → treat as the 4th position = 1.

Question: Unit digit of 3^25?
Cycle for 3: 3, 9, 7, 1 (length 4). 25 mod 4 = 1 → 1st position = 3.

Factorials and trailing zeros

n! = n × (n−1) × … × 2 × 1. By definition, 0! = 1.

Trailing zeros in n!: each trailing zero requires one factor of 10 = 2 × 5. Since factors of 2 are far more abundant than factors of 5, count only the factors of 5 using Legendre's formula:

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

Question: Trailing zeros in 100!?
= ⌊100/5⌋ + ⌊100/25⌋ + ⌊100/125⌋ = 20 + 4 + 0 = 24.

Surds and indices

Definition: A surd is an irrational number expressed using a radical, such as √2 or ³√5, that cannot be simplified to a rational number.

Index laws:

Law	Statement
Product	a^m × a^n = a^(m+n)
Quotient	a^m ÷ a^n = a^(m−n)
Power of a power	(a^m)^n = a^(mn)
Zero exponent	a^0 = 1 (a ≠ 0)
Negative exponent	a^(−n) = 1/a^n
Fractional exponent	a^(1/n) = ⁿ√a

Surd operations:

√a × √b = √(ab)
√a / √b = √(a/b)
(√a)² = a
Rationalising the denominator: multiply numerator and denominator by the conjugate. E.g., 1/(√3+1) × (√3−1)/(√3−1) = (√3−1)/2.

Worked examples

Q1 — LCM/HCF: LCM and HCF of 18 and 24?

Answer: HCF = 6, LCM = 72 (see prime factorisation above).

Q2 — Trailing zeros: How many zeros at the end of 100!?

Answer: 24.

Q3 — Unit digit: Unit digit of 3^25?

Answer: 3 (cycle position 1, since 25 mod 4 = 1).

Q4 — AP from divisibility: Sum of all 2-digit numbers divisible by 7?

List: 14, 21, 28, …, 98 (AP with a = 14, d = 7, last term = 98).
n = (98 − 14)/7 + 1 = 13.
Sum = 13/2 × (14 + 98) = 13 × 56 = 728.

Why it matters: The number system is the bedrock topic for SSC CGL, RRB NTPC, IBPS, CSAT and all school mathematics. Divisibility rules, LCM/HCF, unit digits and trailing zeros appear in nearly every shift of competitive exams. Mastering these is non-negotiable for a high score in quantitative aptitude.

Real-world example: The LCM concept appears directly in everyday scheduling — if Bus A departs every 12 minutes and Bus B every 18 minutes, they leave together every LCM(12, 18) = 36 minutes. That is a standard exam template repeated in dozens of variations.

Common misconception: Students frequently confuse HCF and LCM applications. HCF is used to split (find the largest equal group), LCM is used to combine (find when events coincide). Also, 1 is neither prime nor composite — this trips a remarkable number of students in exam conditions.

:::keypoints Key points

Number sets nest: N ⊂ W ⊂ Z ⊂ Q ⊂ R ⊂ C; each set inherits the previous one's properties.
Rational decimals terminate or repeat; irrational decimals never repeat.
1 is neither prime nor composite; 2 is the only even prime.
HCF uses minimum prime-factor powers; LCM uses maximum powers.
LCM × HCF = product of the two numbers.
Trailing zeros in n! = count of 5-factors via Legendre's formula.
Unit digits of powers follow cycles of length 1, 2, or 4 — find the cycle, then mod the exponent.
Rationalise surds by multiplying by the conjugate of the denominator.
:::
:::memory
"HCF-min, LCM-max" — HCF takes the minimum power of each prime; LCM takes the maximum. "H" for "humble" (smaller), "L" for "large."
:::
:::recap
The number-system hierarchy tells you which operations stay within a set (closure) and which leave it.
Divisibility rules for 2–11 let you test factors mentally without long division.
LCM and HCF have a reciprocal relationship via the golden property; prime factorisation makes both fast to find.
Unit-digit problems reduce to cycle patterns of length ≤ 4; use exponent mod cycle-length.
Trailing zeros in n! count factors of 5 — apply Legendre's formula, not a count of 10s.
:::

Remainders and Divisibility

Remainder Theorem Concepts

Notes

The remainder theorem is to number theory what the percentage formula is to arithmetic — a single compact rule that unlocks an entire category of SSC CGL questions about large powers and complex divisions.

Definition: When a number N is divided by a divisor d, the remainder r satisfies: N = d × q + r, where q is the quotient and 0 ≤ r < d.

Basic Remainder Concepts

Notation: N mod d, or N % d, or "remainder when N ÷ d."

Key observations:

The remainder is always non-negative and strictly less than d
If N is perfectly divisible by d, the remainder is 0
The remainder of N ÷ 1 is always 0 (every integer is divisible by 1)

Real-world example: If 100 students are to sit in rows of 9: 100 = 9 × 11 + 1. The remainder 1 means one student is left over after filling 11 complete rows. This models N = 100, d = 9, q = 11, r = 1.

Modular Arithmetic: The Core Rules

These rules allow you to break a huge computation into smaller, manageable ones.

Rule 1 — Addition mod d:
(a + b) mod d = [(a mod d) + (b mod d)] mod d

Example: (47 + 38) mod 5
= [(47 mod 5) + (38 mod 5)] mod 5
= [2 + 3] mod 5
= 5 mod 5 = 0

Verify: 47 + 38 = 85; 85 ÷ 5 = 17 remainder 0 ✓

Rule 2 — Multiplication mod d:
(a × b) mod d = [(a mod d) × (b mod d)] mod d

Example: (23 × 17) mod 4
= [(23 mod 4) × (17 mod 4)] mod 4
= [3 × 1] mod 4
= 3 mod 4 = 3

Verify: 23 × 17 = 391; 391 = 97 × 4 + 3, remainder 3 ✓

Rule 3 — Powers mod d (derived from Rule 2):
aⁿ mod d = (a mod d)ⁿ mod d

This is the key to all large-power remainder questions.

Why These Rules Work (Intuition)

Think of numbers on a clock-face. A 12-hour clock uses mod 12. If it is 10 o'clock now, and 5 hours pass, the clock shows (10 + 5) mod 12 = 3 — not 15. The "modulus" wraps numbers around just like a circular track.

Real-world example: Weekday calculation. If today is Thursday (day 4, counting Mon=1), what day is it 100 days from now? (4 + 100) mod 7 = 104 mod 7 = 104 − 14×7 = 104 − 98 = 6 → Saturday.

Euler's Totient Theorem

Definition: Euler's Totient function φ(n) counts the number of integers from 1 to n that are co-prime to n.

Euler's Theorem: If gcd(a, n) = 1 (a and n are co-prime), then:
a^φ(n) ≡ 1 (mod n)

This means a^φ(n) leaves remainder 1 when divided by n.

Computing φ(n):

If n = p (prime): φ(p) = p − 1
If n = p^k: φ(p^k) = p^k − p^(k-1) = p^(k-1)(p−1)
If n = p × q (distinct primes): φ(pq) = (p−1)(q−1)

Example: Find remainder when 3^100 is divided by 7.

gcd(3, 7) = 1 ✓
φ(7) = 7 − 1 = 6
3^6 ≡ 1 (mod 7)
100 = 6 × 16 + 4
3^100 = (3^6)^16 × 3^4 ≡ 1^16 × 81 ≡ 81 mod 7
81 = 11 × 7 + 4 → remainder 4

Fermat's Little Theorem

Definition: For a prime p and any integer a where p does not divide a:
a^(p−1) ≡ 1 (mod p)

This is actually a special case of Euler's Theorem (since φ(p) = p−1 for prime p).

Example: Find remainder when 2^10 is divided by 11.

11 is prime; gcd(2, 11) = 1
By Fermat: 2^(11−1) = 2^10 ≡ 1 (mod 11)
Remainder = 1

Why it matters for SSC: When the divisor is prime, Fermat's theorem gives the cycle length instantly without computing powers manually.

Finding Remainders of Large Powers: Step-by-Step Method

For questions like "Find remainder when 7^50 ÷ 5":

Method 1 — Euler/Fermat:

gcd(7, 5) = 1; φ(5) = 4 (since 5 is prime)
7^4 ≡ 1 (mod 5)
50 = 4 × 12 + 2
7^50 = (7^4)^12 × 7^2 ≡ 1^12 × 49 ≡ 49 mod 5
49 = 9 × 5 + 4 → remainder 4

Method 2 — Cyclicity (see next lesson): Build the pattern of remainders manually for 7^1, 7^2, 7^3, 7^4 and find when it repeats.

Both methods must give the same answer.

Special Remainder Shortcuts

Shortcut 1: (N + d) has the same remainder as N when divided by d.
Example: Remainder of 17 ÷ 5 = remainder of (17 − 15) = 2 ÷ 5 = 2.

Shortcut 2: If N ≡ r₁ (mod d) and M ≡ r₂ (mod d), then:

N + M ≡ r₁ + r₂ (mod d)
N × M ≡ r₁ × r₂ (mod d)

Shortcut 3: Negative remainders. Sometimes it is easier to use negative equivalents.

17 mod 9: direct → 17 − 9 = 8 → remainder 8
Alternative: 17 ≡ 17 − 18 = −1 (mod 9). Negative remainder −1 means actual remainder = 9 − 1 = 8 ✓
Negative remainders are powerful for bases just below the modulus.

Worked Example

Question: Find the remainder when 2^100 is divided by 3.

Solution:

Step 1: Note gcd(2, 3) = 1. φ(3) = 2.

Step 2: By Euler: 2^2 ≡ 1 (mod 3). So the cycle length is 2.

Step 3: 100 = 2 × 50, with no remainder. So 2^100 = (2^2)^50 ≡ 1^50 = 1 (mod 3).

Conclusion: Remainder = 1.

Quick verification: 2^1 mod 3 = 2; 2^2 mod 3 = 1; 2^3 mod 3 = 2; 2^4 mod 3 = 1 → cycle is (2, 1). 100 is even → remainder 1 ✓

:::keypoints Key points

Remainder r satisfies N = dq + r where 0 ≤ r < d
(a+b) mod d = [(a mod d) + (b mod d)] mod d — addition distributes over modulus
(a×b) mod d = [(a mod d) × (b mod d)] mod d — multiplication distributes over modulus
Euler: a^φ(n) ≡ 1 mod n when gcd(a,n)=1; φ(prime p) = p−1
Fermat's Little Theorem is Euler's theorem for prime divisors: a^(p−1) ≡ 1 mod p
For large powers: find cycle length from φ(n), then reduce exponent mod cycle
Negative remainders simplify computation when base is just below the divisor
:::

:::memory
"Euler's φ resets to 1" — after φ(n) steps of multiplying a, the remainder returns to 1. Think of it as a timer that resets every φ(n) ticks.
:::

:::recap

Modular arithmetic lets you replace large numbers with their remainders before computing
Euler's theorem: a^φ(n) ≡ 1 (mod n) when a and n are co-prime
Fermat's Little Theorem: a^(p−1) ≡ 1 (mod p) for prime p — use when divisor is prime
To find remainder of a^n: divide n by φ(divisor) to get effective exponent
Negative remainder technique: replace a with (a − d) when a is slightly less than d
Modular addition and multiplication rules allow breaking complex expressions into parts
:::

Cyclicity of Remainders Formula

Formulas

When you need the remainder of 7^100 divided by 5, computing 7^100 directly is impossible by hand — but the cyclicity method turns it into a 4-step problem that takes 20 seconds.

What is Cyclicity?

Definition: Cyclicity (in the context of remainders) is the property that the remainders of successive powers of a number repeat in a fixed cycle when divided by a given divisor.

This happens because there are only a finite number of possible remainders (0 through d−1) when dividing by d. Once a remainder repeats, the entire cycle repeats from that point.

The Cyclicity Method: Step by Step

Step 1: Compute successive powers of the base, taking the remainder at each step.

Step 2: Identify when the remainder first repeats (typically when it returns to the remainder of base^1).

Step 3: Note the cycle length (period of repetition).

Step 4: Divide the exponent by the cycle length. The remainder of this division tells you which position in the cycle the answer falls on.

Detailed Worked Example: 7^50 ÷ 5

Step 1 — Compute remainders of 7^n (mod 5):

7^1 = 7 → 7 mod 5 = 2
7^2 = 49 → 49 mod 5 = 4 (or use 2² mod 5 = 4)
7^3 = 343 → 343 mod 5 = 3 (or 4 × 2 mod 5 = 8 mod 5 = 3)
7^4 = 2401 → 2401 mod 5 = 1 (or 3 × 2 mod 5 = 6 mod 5 = 1)
7^5 → 1 × 2 mod 5 = 2 — cycle restarts!

Cycle: 2, 4, 3, 1 — length 4

Step 2 — Find position of 7^50 in cycle:
50 mod 4 = 2 (since 50 = 4 × 12 + 2)

Step 3 — The answer is the 2nd value in the cycle: 4

Conclusion: Remainder of 7^50 ÷ 5 = 4

Quick shortcut check: 7 ≡ 2 (mod 5), so 7^50 ≡ 2^50 (mod 5). 50 mod 4 = 2 → 2^2 = 4. ✓

Why the Cycle Starts Again

The reason cycles are guaranteed: once a remainder r₀ appears twice, the next remainders are determined only by r₀ (via the multiplication rule), so the sequence must repeat exactly.

The cycle length divides φ(d) (Euler's totient). This is why Euler's Theorem guarantees a cycle of length at most φ(d).

Common Cycle Lengths

:::compare Cycle Lengths for Common Bases mod 10 (Last Digit Patterns)

Base	Cycle	Period
2	2,4,8,6	4
3	3,9,7,1	4
7	7,9,3,1	4
8	8,4,2,6	4
4	4,6	2
9	9,1	2
1	1	1
5	5	1
6	6	1
0	0	1
:::

Real-world application: The last-digit cycle is cyclicity mod 10. Finding the last digit of 3^47: cycle is (3,9,7,1), period 4. 47 mod 4 = 3. Third element in cycle = 7. The last digit of 3^47 is 7.

Wilson's Theorem

Statement: For any prime number p:
(p − 1)! ≡ −1 (mod p)

Equivalently: (p−1)! leaves remainder (p−1) when divided by p.

Example: p = 7. Wilson says 6! ≡ −1 ≡ 6 (mod 7).
Verify: 6! = 720. 720 ÷ 7 = 102 remainder 6. ✓

SSC Application:
Find remainder when 6! is divided by 7.
Answer by Wilson: 720 mod 7 = 6 (i.e., 7−1).

Find remainder when 10! is divided by 11.
11 is prime; Wilson gives 10! ≡ −1 ≡ 10 (mod 11). Remainder = 10.

Converse: If (n−1)! ≡ −1 (mod n), then n is prime. This can be used to verify primality but is impractical for large n by hand.

Chinese Remainder Theorem (CRT) — Concept

Definition: The Chinese Remainder Theorem states that if you know the remainders of a number N when divided by several pairwise co-prime moduli, you can uniquely determine N (modulo the product of those moduli).

Practical SSC Form:

"A number N leaves remainder 3 when divided by 5, and remainder 2 when divided by 7. Find the smallest such N."

Method:

N ≡ 3 (mod 5): N can be 3, 8, 13, 18, 23, 28, 33, 38…
N ≡ 2 (mod 7): N can be 2, 9, 16, 23, 30, 37… wait, let's list mod 7 residue 2: 2, 9, 16, 23, 30, 37…
Check N = 23: 23 mod 5 = 3 ✓; 23 mod 7 = 2 ✓. N = 23.

The general solution: N = 23 + k × LCM(5,7) = 23 + 35k. Smallest positive N = 23.

Worked Example: Full Cyclicity Problem

Question: Find the remainder when 13^51 is divided by 7.

Solution:

Step 1: Reduce base first. 13 mod 7 = 6 (since 13 = 7×1 + 6). So 13^51 ≡ 6^51 (mod 7).

Step 2: Note 6 ≡ −1 (mod 7). So 6^51 ≡ (−1)^51 = −1 ≡ 6 (mod 7).

Conclusion: Remainder = 6.

(This shows the power of recognising a base that is "−1" relative to the modulus.)

Worked Example: Wilson's Theorem Application

Question: Find the remainder when 14! is divided by 17.

Solution:

Step 1: 17 is prime; Wilson says 16! ≡ −1 ≡ 16 (mod 17).

Step 2: 16! = 16 × 15 × 14!. So 16 × 15 × (14!) ≡ 16 (mod 17).

Step 3: 16 ≡ −1 (mod 17) and 15 ≡ −2 (mod 17).
(−1)(−2)(14!) ≡ 16 ≡ −1 (mod 17)
2 × 14! ≡ −1 (mod 17)
14! ≡ −1 × (2^−1 mod 17) (mod 17)

Step 4: Find 2^−1 mod 17: 2 × 9 = 18 ≡ 1 (mod 17), so 2^−1 = 9.
14! ≡ −9 ≡ 17 − 9 = 8 (mod 17).

Conclusion: Remainder = 8.

:::keypoints Key points

Cyclicity: remainders of successive powers repeat in a fixed cycle of period ≤ φ(d)
Method: compute r¹, r², r³… until cycle repeats; find exponent mod cycle length
For mod 10 (last digit): bases 2,3,7,8 have period 4; bases 4,9 have period 2; bases 1,5,6 have period 1
Wilson's Theorem: for prime p, (p−1)! ≡ −1 (mod p)
CRT allows finding a number from its remainders under co-prime divisors
Base ≡ −1 (mod d) trick: power is −1 if exponent is odd, +1 if exponent is even
Always reduce the base modulo d first before starting the cycle computation
:::

:::memory
"Cycle, Reduce, Position" — Three steps: find the cycle, reduce the exponent (mod cycle length), pick the position. For Wilson: "(Prime−1)! = −1" — the factorial just misses a multiple.
:::

:::recap

Cyclicity method: map successive powers to remainders, spot the repeating cycle
Cycle length divides φ(d); for prime d, cycle length divides (d−1)
Reduce large base to its remainder first; this simplifies all subsequent calculations
Wilson's Theorem: (p−1)! ≡ p−1 (mod p) for prime p — useful for factorial remainders
CRT: combine congruences under co-prime moduli using LCM-based enumeration
"Base = d−1" shortcut: any (d−1)^n ≡ (−1)^n mod d
:::

Worked Example: Finding Remainders

Worked example

The binomial identity (a + 1)^n = multiple of a + 1 is one of the most elegant shortcuts in number theory — once you see it, an entire class of large-power remainder questions collapses into a single step.

The Core Principle: Binomial Expansion Shortcut

Why it works: By the Binomial Theorem, (kd + r)^n expands as:
(kd)^n + n(kd)^(n−1)·r + … + n·(kd)·r^(n−1) + r^n

Every term except the last contains kd as a factor, so every term except the last is divisible by d. This means:
(kd + r)^n ≡ r^n (mod d)

Special case r = 1:
(kd + 1)^n ≡ 1^n = 1 (mod d) — always remainder 1, regardless of n.

Special case r = −1 (i.e., kd − 1):
(kd − 1)^n ≡ (−1)^n (mod d)

If n is even: remainder = 1
If n is odd: remainder = d − 1

The Primary Worked Example: 17^35 ÷ 16

Question: Find the remainder when 17^35 is divided by 16.

Solution:

Step 1: Notice that 17 = 16 + 1. So here d = 16, k = 1, r = 1, n = 35.

Step 2: By the binomial rule: (16 + 1)^35 ≡ 1^35 = 1 (mod 16).

Conclusion: Remainder = 1.

Intuitive check: 17 ≡ 1 (mod 16). So 17^35 ≡ 1^35 = 1 (mod 16). ✓

This takes under 10 seconds once you spot the pattern.

The Companion Rule: (kd − 1)^n

Rule: (kd − 1)^n mod d:

n even → remainder = 1
n odd → remainder = d − 1

Derivation: kd − 1 ≡ −1 (mod d). So (kd−1)^n ≡ (−1)^n (mod d). When n is even, (−1)^n = 1 → remainder 1. When n is odd, (−1)^n = −1 ≡ d−1 (mod d) → remainder d−1.

Extended Examples

Example A: Find the remainder when 19^41 is divided by 18.

19 = 18 + 1, so 19 ≡ 1 (mod 18)
19^41 ≡ 1^41 = 1 (mod 18)
Remainder = 1

Example B: Find the remainder when 35^99 is divided by 36.

35 = 36 − 1, so 35 ≡ −1 (mod 36)
99 is odd → (−1)^99 = −1 ≡ 35 (mod 36)
Remainder = 35

Example C: Find the remainder when 35^100 is divided by 36.

35 ≡ −1 (mod 36); 100 is even → (−1)^100 = 1
Remainder = 1

Example D: Find the remainder when 31^26 is divided by 32.

31 = 32 − 1 ≡ −1 (mod 32); 26 is even
Remainder = 1

More General Binomial Reduction

When r ≠ 1 or −1, the binomial rule still reduces the problem:

Example E: Find the remainder when 23^3 is divided by 10.

23 = 20 + 3. So 23^3 ≡ 3^3 = 27 ≡ 7 (mod 10).
Remainder = 7

Example F: Find the remainder when 52^10 is divided by 7.

52 = 49 + 3 = 7×7 + 3. So 52 ≡ 3 (mod 7).
52^10 ≡ 3^10 (mod 7).
By Fermat: 3^6 ≡ 1 (mod 7) (since 7 is prime).
3^10 = 3^6 × 3^4 ≡ 1 × 81 mod 7 = 81 mod 7 = 4 (81 = 11×7 + 4).
Remainder = 4

Combining Techniques: A Harder Example

Question: Find the remainder when 2^200 is divided by 15.

Solution:

Step 1: Note gcd(2, 15) = 1. φ(15) = φ(3)×φ(5) = 2×4 = 8.
By Euler: 2^8 ≡ 1 (mod 15).

Step 2: 200 = 8 × 25. So 2^200 = (2^8)^25 ≡ 1^25 = 1 (mod 15).

Conclusion: Remainder = 1.

Quick verify: 2^4 = 16 ≡ 1 (mod 15). So cycle length is actually 4 (not 8 — Euler gives maximum, actual can be shorter). 200 = 4 × 50 → 2^200 ≡ 1^50 = 1 (mod 15). ✓

Summary of Pattern-Based Shortcuts

:::compare Remainder Patterns

Form	Condition	Remainder mod d
(kd + 1)^n	Any n	1
(kd − 1)^n	n even	1
(kd − 1)^n	n odd	d − 1
(kd + r)^n	General	r^n mod d
a^n, a ≡ 1 (mod d)	Any n	1
a^n, a ≡ −1 (mod d)	n even	1
a^n, a ≡ −1 (mod d)	n odd	d − 1
:::

Worked Practice Set

Question 1: Remainder when 101^99 ÷ 100.

101 = 100 + 1 → remainder = 1

Question 2: Remainder when 99^99 ÷ 100.

99 = 100 − 1; 99 is odd → remainder = 99

Question 3: Remainder when 99^100 ÷ 100.

99 = 100 − 1; 100 is even → remainder = 1

Question 4: Remainder when 37^2 ÷ 13.

37 = 3×13 − 2 → 37 ≡ −2 (mod 13)
37^2 ≡ (−2)^2 = 4 → remainder = 4

:::keypoints Key points

(kd + 1)^n always leaves remainder 1 when divided by d
(kd − 1)^n leaves remainder 1 (even n) or d−1 (odd n) when divided by d
General binomial rule: (kd + r)^n ≡ r^n (mod d)
Spot if base = d+1 or d−1 first — this is the fastest possible path to the answer
When the base is more general, reduce it (mod d) first, then apply Euler/Fermat or cyclicity
Negative remainder −1 mod d = d−1; use this to simplify odd-power cases
:::

:::memory
"Plus-one never tires — always leaves 1. Minus-one plays odds and evens — odd power gives (d−1), even power gives 1."
:::

:::recap

The key identity: base = (multiple of d) + 1 → remainder is always 1
Base = (multiple of d) − 1 → remainder alternates: 1 for even exponent, d−1 for odd exponent
These follow directly from binomial expansion — all terms divisible by d vanish
For general bases, first reduce mod d, then apply cycle/Euler/Fermat
Practice recognising (d+1) and (d−1) forms instantly — they appear in 40%+ of SSC remainder questions
:::

Simplification and BODMAS

BODMAS Rule and Order of Operations

Notes

Three or four marks in SSC CGL Tier 1 come straight from simplification — and every single one of them is lost or won on one rule: the correct order of operations.

Definition: BODMAS is the universally accepted precedence order for arithmetic operations: Brackets → Orders (powers and roots) → Division → Multiplication → Addition → Subtraction.

Breaking Down BODMAS

B — Brackets

Four types of brackets, resolved from innermost to outermost:

Level	Symbol	Name	Resolve order
1 (innermost)	—	Vinculum (overbar)	First
2	( )	Round bracket	Second
3	{ }	Curly bracket	Third
4 (outermost)	[ ]	Square bracket	Last

Example: [ {(3 + 2) × 4} − 10 ]

Step 1 — Round: (3 + 2) = 5
Step 2 — Curly: {5 × 4} = 20
Step 3 — Square: [20 − 10] = 10

Vinculum example: 3 + 4̄+6̄ × 2 (where the bar covers 4+6)

Vinculum first: 4+6 = 10 (the bar acts like hidden brackets)
Then: 3 + 10 × 2 = 3 + 20 = 23

O — Orders (Powers and Roots)

After all brackets are cleared, evaluate exponents and roots.

Example: 2 + 3² − √16 + 5
= 2 + 9 − 4 + 5 (evaluate 3² = 9 and √16 = 4 first)
= 12

D and M — Division and Multiplication (Equal Priority)

Critical rule: Division and Multiplication have the same priority — do them left to right as they appear.

The most common SSC mistake:

20 ÷ 4 × 5
Wrong: 20 ÷ (4 × 5) = 20 ÷ 20 = 1 ✗
Correct: (20 ÷ 4) × 5 = 5 × 5 = 25 ✓

Another example: 48 ÷ 6 ÷ 2

Left to right: (48 ÷ 6) ÷ 2 = 8 ÷ 2 = 4 ✓
NOT 48 ÷ (6 ÷ 2) = 48 ÷ 3 = 16 ✗

A and S — Addition and Subtraction (Equal Priority)

Similarly, Addition and Subtraction have equal priority — go left to right.

Example: 10 − 3 + 5
= (10 − 3) + 5 = 7 + 5 = 12 ✓
NOT 10 − (3 + 5) = 10 − 8 = 2 ✗

Why "BODMAS" Works as a Mnemonic

In India, BODMAS is used universally. Some countries use PEMDAS or BIDMAS — same order, different mnemonics:

P = Parentheses = B (Brackets)
E = Exponents = O (Orders)
MD = Multiplication/Division (left to right)
AS = Addition/Subtraction (left to right)

Real-world example: Cooking recipe proportions. If a recipe calls for (2 + 3) × 4 cups of flour: you add first (inside brackets), then multiply. If you ignore brackets and do 2 + 3 × 4 = 2 + 12 = 14 instead of (2+3) × 4 = 20, you ruin the recipe.

Common SSC CGL Question Types

Type 1 — Pure Simplification

Question: Simplify: 18 − 3 × 4 + 12 ÷ 3

Solution:
Step 1 (O): None.
Step 2 (D/M left-to-right): 3 × 4 = 12; 12 ÷ 3 = 4.
Expression becomes: 18 − 12 + 4
Step 3 (A/S left-to-right): (18 − 12) + 4 = 6 + 4 = 10

Type 2 — Nested Brackets

Question: Simplify: [5 + {3 × (8 − 4) ÷ 6}]

Solution:
Step 1 — Round bracket: (8 − 4) = 4
Step 2 — Curly bracket (left to right): 3 × 4 = 12; 12 ÷ 6 = 2. {3 × 4 ÷ 6} = 2
Step 3 — Square bracket: [5 + 2] = 7

Type 3 — Vinculum (bar notation)

Question: Simplify: 2 + 3 × 4̄+2̄ (bar over "4+2")

Solution:
Step 1 — Vinculum: 4 + 2 = 6
Step 2 — Multiplication: 3 × 6 = 18
Step 3 — Addition: 2 + 18 = 20

Type 4 — Mixed with Powers

Question: Simplify: 4² − 3 × (6 − 2) + 8 ÷ 2

Solution:
Step 1 — Brackets: (6 − 2) = 4
Step 2 — Orders: 4² = 16
Step 3 — D/M left-to-right: 3 × 4 = 12; 8 ÷ 2 = 4. Expression: 16 − 12 + 4
Step 4 — A/S left-to-right: (16 − 12) + 4 = 4 + 4 = 8

The Five Most Common Mistakes

Mistake 1 — Treating D and M as having different priorities:
24 ÷ 3 × 2 = 8 × 2 = 16 (correct). NOT 24 ÷ 6 = 4.

Mistake 2 — Same for A and S:
10 − 4 + 3 = 6 + 3 = 9 (correct). NOT 10 − 7 = 3.

Mistake 3 — Ignoring vinculum:
Students treat bar-notation the same as multiplication without parentheses.

Mistake 4 — Applying operations right-to-left:
Division especially feels "natural" right-to-left, but rules require left-to-right.

Mistake 5 — Skipping the bracket hierarchy:
Solving square brackets before inner round brackets.

Full Worked Example (SSC CGL Style)

Question: Simplify: [3 + {5 × (12 − 8)² ÷ 20} − 2]

Solution:

Step 1 — Innermost bracket (round): (12 − 8) = 4

Step 2 — Orders: 4² = 16

Step 3 — Curly bracket (D/M left-to-right): 5 × 16 = 80; 80 ÷ 20 = 4. So {5 × (12−8)² ÷ 20} = 4

Step 4 — Square bracket: [3 + 4 − 2] = [7 − 2] = 5

SSC CGL Examination Insight

SSC CGL Tier I typically includes 3–5 simplification questions. They test:

Pure BODMAS order (no tricks)
Bracket hierarchy
Fraction simplification with BODMAS
Mixed operations with integers and decimals

Time target: 30–45 seconds per simplification question. If you find yourself spending more time, practice until the order of operations is automatic.

:::keypoints Key points

BODMAS: Brackets → Orders → Division → Multiplication → Addition → Subtraction
Four bracket types solved innermost-first: vinculum → ( ) → { } → [ ]
Division and Multiplication have EQUAL priority — solve strictly left to right
Addition and Subtraction have EQUAL priority — solve strictly left to right
The most common error: doing right multiplication before left division in D/M step
In SSC CGL, 3–5 marks per paper come directly from simplification
:::

:::memory
"Big Owls Don't Make Any Sound" → B-O-D-M-A-S. For brackets: "Vicious Round Curly Square" (Vinculum, Round, Curly, Square — inner to outer).
:::

:::recap

BODMAS is the non-negotiable precedence order for all arithmetic expressions
Always clear the innermost bracket type first; vinculum (overbar) is innermost of all
D and M: no priority difference — work left to right through the expression
A and S: no priority difference — work left to right through the expression
Practice the classic trap: 20 ÷ 4 × 5 = 25 (not 1)
Three to four simplification questions appear in SSC CGL Tier I — all are winnable with this one rule
:::

Useful Simplification Formulas

Formulas

Exam halls have claimed countless marks from students who knew the right method but couldn't execute fast enough — mastering these simplification formulas turns a 3-minute slog into a 30-second win.

Algebraic Identities: Your Speed-Calculation Arsenal

Algebraic identities are equations that hold true for all values of the variables. Unlike equations that you solve, identities are tools you apply — they let you expand, factor, and rearrange expressions without grinding through long multiplication.

The Standard Expansion Identities

Definition: An algebraic identity is an equality that is satisfied by every value of the variable(s) involved.

The six identities you must memorise cold:

Identity	Form	When to use
(a + b)² = a² + 2ab + b²	Perfect square sum	Expand squares of binomials
(a − b)² = a² − 2ab + b²	Perfect square difference	Same, with subtraction
a² − b² = (a + b)(a − b)	Difference of squares	Factor or evaluate quickly
(a + b)³ = a³ + 3a²b + 3ab² + b³	Cube expansion	Expand cubed binomials
a³ + b³ = (a + b)(a² − ab + b²)	Sum of cubes	Factor sum-of-cubes expressions
a³ − b³ = (a − b)(a² + ab + b²)	Difference of cubes	Factor difference-of-cubes

Why it matters: SSC CGL, RRB NTPC, and banking exams frequently ask you to find the value of expressions like a² + b² given a + b and ab. Without these identities you'd need to find a and b separately — with them you can compute the answer in one line.

Using Identities Backwards (Factoring Direction)

The real power comes from using identities in reverse. Consider 9x² − 25. Recognise this as (3x)² − (5)², which immediately factors as (3x + 5)(3x − 5). This "reverse application" is the key to simplifying complex fractions and expressions.

Real-world example (exam-style): If a + b = 7 and ab = 12, find a² + b².

Using (a + b)² = a² + 2ab + b²:
49 = a² + 24 + b²
a² + b² = 25

No need to solve for a and b individually.

Fraction Shortcuts

Division of Fractions

Definition: To divide one fraction by another, multiply the first fraction by the reciprocal of the second.

a/b ÷ c/d = a/b × d/c = ad/bc

Common misconception: Many students try to divide numerators and denominators separately — (a ÷ c)/(b ÷ d) — which gives the wrong answer unless a/c and b/d simplify cleanly. Always flip-and-multiply.

Question: Simplify (3/5) ÷ (9/10).
Solution: (3/5) × (10/9) = 30/45 = 2/3.

Mixed Fractions

Definition: A mixed fraction (or mixed number) combines a whole number and a proper fraction: 3½ means 3 + ½.

Conversion rule: 3½ = (3 × 2 + 1)/2 = 7/2.

General rule: a(b/c) = (ac + b)/c.

Why it matters: Exam questions often present data in mixed fractions. Converting incorrectly is one of the most common sources of careless errors.

Surds: Rules for Working with Square Roots

Definition: A surd is an irrational root that cannot be simplified to a rational number — for example, √2, √3, √7.

The Three Surd Laws

Product rule: √a × √b = √(ab) (only valid when a ≥ 0, b ≥ 0)
Quotient rule: √a / √b = √(a/b)
Conjugate product: (√a + √b)(√a − √b) = a − b

The third rule is especially powerful for rationalising denominators — a common exam task.

Worked Example:

Question: Rationalise 1/(√5 + √2).

Solution:
Step 1: Multiply numerator and denominator by the conjugate (√5 − √2).
Step 2: Denominator = (√5)² − (√2)² = 5 − 2 = 3.
Step 3: Numerator = (√5 − √2).
Conclusion: Result = (√5 − √2)/3.

Common misconception: √(a + b) ≠ √a + √b. This is wrong for almost all values. For example, √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7. The product rule works; the addition rule does not.

Quick Decimal Equivalents to Memorise

Fraction arithmetic becomes much faster when you have common conversions memorised. These appear directly in options or are needed to compare quantities:

Fraction	Decimal	Memory hook
1/2	0.5	Base reference
1/3	0.333…	Repeating 3s
1/4	0.25	Half of half
1/5	0.2	One-fifth
1/6	0.1667	1/2 of 1/3
1/7	0.1429	7 goes ~7 times into 1
1/8	0.125	Half of 0.25
1/9	0.1111	Repeating 1s

Why it matters: Comparing options like "is 5/7 or 8/11 larger?" becomes immediate if you convert: 5/7 ≈ 0.714, 8/11 ≈ 0.727 → 8/11 is larger. No cross-multiplication needed.

Bringing It Together: A Multi-Step Simplification

Question: Evaluate [(√5 + √3)² − (√5 − √3)²] / (2√15).

Solution:
Step 1: Expand using (a+b)² and (a−b)²:
(√5 + √3)² = 5 + 2√15 + 3 = 8 + 2√15
(√5 − √3)² = 5 − 2√15 + 3 = 8 − 2√15

Step 2: Subtract: (8 + 2√15) − (8 − 2√15) = 4√15.

Step 3: Divide: 4√15 / 2√15 = 2.

Conclusion: The answer is 2.

Notice how using identities directly avoided any need to compute √5 or √3 individually.

:::compare Identity vs. Direct Expansion

Approach	Steps for (a+b)²	Time cost
Identity	1 line: a² + 2ab + b²	~5 seconds
Direct multiply	(a+b)(a+b), 4 partial products	~30 seconds
Error risk	Very low	Medium
:::

:::keypoints Key points

(a+b)² and (a−b)² differ only in the sign of the middle term 2ab.
a²−b² = (a+b)(a−b) is the fastest way to evaluate near-square products (e.g., 99×101 = 100²−1 = 9999).
Division of fractions: flip the second fraction and multiply.
Mixed fraction a(b/c) converts to (ac+b)/c — always convert before computing.
√a × √b = √(ab) works; √(a+b) ≠ √a + √b — this is a critical misconception.
Rationalise denominators using the conjugate surd: (√a+√b)(√a−√b) = a−b.
Decimal equivalents for 1/6, 1/7, 1/8, 1/9 appear often in comparison questions.
Identities work backwards (factoring) as well as forwards (expansion).
:::

:::memory
"Some People Are Dumb, And Can't Simplify Fractions" → Square, Product, And Difference, Add Cubes, Sum-of-cubes, Factor — the six identities in order. For surds, think "Multiply but never Add under the root."
:::

:::recap

Six core identities cover squares, cubes, and difference of squares — memorise both the expanded and factored forms.
(a+b)² = a² + 2ab + b² and (a−b)² = a² − 2ab + b²; the only difference is the sign of 2ab.
Fraction division: multiply by the reciprocal (flip-and-multiply).
Mixed number a(b/c) → (ac+b)/c before any calculation.
Surd product rule holds: √a × √b = √(ab); sum rule does NOT: √(a+b) ≠ √a + √b.
Know 1/6, 1/7, 1/8, 1/9 as decimals to speed up comparison problems.
:::

Solved Example: BODMAS Application

Worked example

One wrongly-placed bracket can cost you the entire mark — BODMAS is not just a rule to recall but a discipline to apply step by step, every single time.

Definition: BODMAS (also written BIDMAS or PEMDAS in other syllabi) is the standard order of operations: Brackets, Of (powers/exponents), Division, Multiplication, Addition, Subtraction. Operations of equal precedence (D & M, or A & S) are evaluated left to right.

Understanding the Bracket Hierarchy

When expressions contain nested brackets, you work from the innermost bracket outward:

Bracket Type	Symbol	Order of evaluation
Round (parentheses)	( )	First (innermost)
Curly (braces)	{ }	Second
Square	[ ]	Third (outermost)

Why it matters: Indian competitive exams — SSC CGL, SSC CHSL, RRB NTPC — consistently include 2–3 BODMAS questions per Quantitative Aptitude section. A single sign error typically shifts the answer by exactly the value that eliminates your option, making it easy to "trap" hasty solvers.

Full Worked Example 1 (Nested Brackets)

Question: Simplify 18 − [6 − {4 + (8 − 3)}].

Solution:

Step 1: Innermost bracket ( ): 8 − 3 = 5.
Expression becomes: 18 − [6 − {4 + 5}]

Step 2: Curly bracket { }: 4 + 5 = 9.
Expression becomes: 18 − [6 − 9]

Step 3: Square bracket [ ]: 6 − 9 = −3.
Expression becomes: 18 − (−3)

Step 4: Subtracting a negative equals adding: 18 + 3 = 21.

Conclusion: The answer is 21.

Common misconception corrected: "Subtracting a negative" trips up the majority of students. When you see − (−3), think of it as removing a debt of 3 — you gain 3. Algebraically: −(−3) = +3. This single rule accounts for a large fraction of wrong answers in BODMAS questions.

Full Worked Example 2 (Order of Operations, No Extra Brackets)

Question: Evaluate 5 + 3 × 4 − 2.

Solution:

Step 1: Multiplication before Addition/Subtraction: 3 × 4 = 12.
Expression becomes: 5 + 12 − 2.

Step 2: Left to right: 5 + 12 = 17, then 17 − 2 = 15.

Conclusion: The answer is 15 — NOT 30.

Common misconception corrected: A very common error is to read left-to-right naively: 5 + 3 = 8, then 8 × 4 = 32, then 32 − 2 = 30. This is wrong because multiplication must be performed before addition regardless of position. BODMAS applies universally.

Worked Example 3 (Mixing All Operations)

Question: Simplify 36 ÷ 6 + 2² × 3 − (8 − 5).

Solution:

Step 1: Brackets first: (8 − 5) = 3.
Expression: 36 ÷ 6 + 2² × 3 − 3.

Step 2: "O" — orders/exponents: 2² = 4.
Expression: 36 ÷ 6 + 4 × 3 − 3.

Step 3: Division and Multiplication (left to right):
36 ÷ 6 = 6; 4 × 3 = 12.
Expression: 6 + 12 − 3.

Step 4: Addition and Subtraction (left to right):
6 + 12 = 18; 18 − 3 = 15.

Conclusion: The answer is 15.

Worked Example 4 (The "Of" Trap)

Question: Find: 1/2 of 48 + 6 × 2 − 10.

Solution:

Step 1: "Of" acts like multiplication and comes after brackets, same rank as ×:
1/2 of 48 = 24; separately 6 × 2 = 12.
Expression: 24 + 12 − 10.

Step 2: Left to right: 24 + 12 = 36; 36 − 10 = 26.

Conclusion: The answer is 26.

Why it matters: "Of" in problems like "one-third of 90" is multiplication in disguise. Students who treat it as addition last are consistently wrong.

A Framework for Error-Free BODMAS

Use this checklist mentally before writing each step:

Scan the entire expression — identify all bracket layers.
Peel from inside out: ( ) first, then { }, then [ ].
Check for exponents immediately after all brackets are resolved.
Do all × and ÷ left-to-right in one pass.
Do all + and − left-to-right in one pass.
Verify sign whenever subtracting — a −(−n) is always +n.

:::compare Common BODMAS Errors vs. Correct Approach

Mistake	Incorrect Result	Correct Approach
Left-to-right without precedence: 5+3×4	(5+3)×4 = 32	5+(3×4) = 17
Forgetting −(−n) = +n	18−(−3) = 15	18−(−3) = 21
Ignoring exponent before ×	2²×3 treated as (2×2×3) by guessing	Compute 2²=4 first, then ×3
Wrong bracket order	Solving [ ] before ( )	Always ( ) → { } → [ ]
:::

:::keypoints Key points

BODMAS order: Brackets → Of/Orders → Division → Multiplication → Addition → Subtraction.
Nested brackets are solved inside-out: ( ) before { } before [ ].
Division and multiplication have equal precedence — work left-to-right when both appear.
Same rule applies for addition and subtraction — left-to-right.
Subtracting a negative is addition: a − (−b) = a + b.
"Of" means multiply, and it comes before addition/subtraction.
Exponents (powers, roots) are evaluated before multiplication.
Always rewrite the expression after each step to avoid carry-forward errors.
:::

:::memory
"Big Orange Dogs Must Always Sit" → Brackets, Orders, Division, Multiplication, Addition, Subtraction. And for the most common trap: "Minus Minus Makes a Plus" (−)(−) = (+).
:::

:::recap

BODMAS governs the exact sequence of arithmetic operations — skipping steps causes systematic errors.
Always work from the innermost bracket outward: ( ) then { } then [ ].
Multiplication and division outrank addition and subtraction; when mixed, go left to right.
The most common error is ignoring that −(−n) = +n; always check sign changes.
After brackets, handle exponents/powers before any multiplication.
Write out each intermediate expression explicitly — never skip steps under exam pressure.
:::

Number Patterns and Series

Important Number Sequences and Formulas

Formulas

The difference between a student who scores 45/50 in Quantitative Aptitude and one who scores 35/50 is often not intelligence — it is whether they have these seven series formulas instant-recalled versus slowly derived.

Definition: A number sequence (or series) is an ordered list of numbers following a specific rule. A series is the sum of the terms of a sequence.

Why Series Formulas Appear in Every Exam

SSC CGL, SSC CHSL, RRB NTPC, and banking exams test series in two ways:

Direct computation — "Find the sum of the first 50 odd numbers."
Missing term — "Find the missing term in: 2, 6, 12, 20, ?, 42."

Formula knowledge collapses a 2-minute calculation to a 10-second substitution.

Core Summation Formulas

Formula 1: Sum of First n Natural Numbers

S = n(n + 1) / 2

Natural numbers are 1, 2, 3, 4, … This formula is attributed to Gauss, who as a child reportedly summed 1 to 100 by pairing: (1+100), (2+99), ... = 50 pairs of 101 → 5050.

Verification: n = 4 → 1+2+3+4 = 10; formula: 4×5/2 = 10. ✓

Formula 2: Sum of Squares of First n Natural Numbers

S = n(n + 1)(2n + 1) / 6

This counts the total number of unit squares in layered arrangements — relevant in geometry and combinatorics, but tested purely computationally in exams.

Verification: n = 3 → 1+4+9 = 14; formula: 3×4×7/6 = 14. ✓

Formula 3: Sum of Cubes of First n Natural Numbers

S = [n(n + 1) / 2]²

Notice this equals the square of Formula 1. This elegant identity means: (sum of first n natural numbers)² = sum of their cubes.

Verification: n = 3 → 1+8+27 = 36; formula: [3×4/2]² = 6² = 36. ✓

Formula 4: Sum of First n Even Numbers

S = n(n + 1)

The first n even numbers are 2, 4, 6, ..., 2n.

Verification: n = 4 → 2+4+6+8 = 20; formula: 4×5 = 20. ✓

Formula 5: Sum of First n Odd Numbers

S = n²

The first n odd numbers are 1, 3, 5, ..., (2n−1).

Verification: n = 4 → 1+3+5+7 = 16; formula: 4² = 16. ✓

Real-world Indian example: The number of bricks in a square staircase arrangement — 1 on the top step, 3 on the next, 5 on the next — equals n² for n steps. This is directly the sum of first n odd numbers.

Formula 6: Sum of an Arithmetic Series (General)

S = (n/2) × (first term + last term)

or equivalently S = (n/2) × [2a + (n−1)d], where a = first term, d = common difference.

Definition: An arithmetic progression (AP) is a sequence where consecutive terms differ by a constant value called the common difference (d).

Worked Example:

Question: Find the sum of the AP: 3, 7, 11, ..., 99.

Solution:
Step 1: Identify a = 3, d = 4.
Step 2: Find n. Last term = a + (n−1)d → 99 = 3 + (n−1)×4 → n−1 = 24 → n = 25.
Step 3: S = (25/2) × (3 + 99) = (25/2) × 102 = 25 × 51 = 1275.
Conclusion: Sum = 1275.

Identifying Missing Terms in Number Series

When asked to find a missing term, you must first decode the pattern. Common pattern types:

Pattern Type	Example	Rule
Constant difference (AP)	3, 7, 11, ?, 19	+4 each time
Constant ratio (GP)	2, 6, 18, ?, 162	×3 each time
Increasing difference	1, 2, 4, 7, 11, ?	Differences: +1, +2, +3, +4...
Perfect squares	1, 4, 9, 16, ?, 36	n² sequence
Perfect cubes	1, 8, 27, ?, 125	n³ sequence
Prime numbers	2, 3, 5, 7, ?, 13	Consecutive primes
Fibonacci-type	1, 1, 2, 3, 5, ?	Each term = sum of two before

Worked Example (Missing Term):

Question: Find the missing term: 2, 6, 12, 20, ?, 42.

Solution:
Step 1: Write differences: 6−2=4, 12−6=6, 20−12=8, ?−20=?, 42−?=?
Step 2: Differences are 4, 6, 8 — increasing by 2 each time. Next difference = 10.
Step 3: Missing term = 20 + 10 = 30. Check: 42 − 30 = 12 ✓ (next difference in sequence).
Conclusion: Missing term = 30.

Alternative recognition: The series is n(n+1): 1×2=2, 2×3=6, 3×4=12, 4×5=20, 5×6=30, 6×7=42. ✓

Applying the "Sum of Odd Numbers" Formula

Question: Find the sum 1 + 3 + 5 + 7 + ... + 99.

Solution:
Step 1: Count terms. The kth odd number = 2k−1. Set 2k−1=99 → k=50. So n=50.
Step 2: Sum = n² = 50² = 2500.
Conclusion: Sum = 2500.

:::compare Summation Formulas Side by Side

Series	Formula	Example (n=5)
1+2+3+…+n	n(n+1)/2	15
1²+2²+…+n²	n(n+1)(2n+1)/6	55
1³+2³+…+n³	[n(n+1)/2]²	225
2+4+6+…+2n	n(n+1)	30
1+3+5+…+(2n−1)	n²	25
:::

:::keypoints Key points

Sum of first n natural numbers: n(n+1)/2 — the Gauss formula.
Sum of first n squares: n(n+1)(2n+1)/6 — includes 2n+1 factor.
Sum of first n cubes = square of sum of first n naturals: [n(n+1)/2]².
Sum of first n even numbers: n(n+1) — exactly double the natural numbers formula.
Sum of first n odd numbers: n² — clean, elegant, fast.
For any AP: S = (n/2)(first + last) — only requires knowing endpoints and count.
To find n for a series ending at term L: n = (L − a)/d + 1.
Pattern identification in missing-term problems: check differences, ratios, and recognise square/cube/prime sequences.
:::

:::memory
"Never Square Cubes Evenly Or Awkwardly" → Natural, Squares, Cubes, Even, Odd, AP — the six formula types. For values: think n/2 × (first+last) works for ALL arithmetic sums — the special formulas are just fast versions of this.
:::

:::recap

Five key formulas cover natural numbers, squares, cubes, even numbers, and odd numbers — each worth knowing as a reflex.
The sum of first n odd numbers equals n² — elegant and instantly testable.
Sum of first n cubes equals the square of the sum of first n naturals — a powerful identity.
For general APs, use S = (n/2)(first + last) or S = (n/2)[2a + (n−1)d].
Missing-term problems require identifying the underlying rule: AP (constant difference), GP (constant ratio), or a recognisable sequence like primes, squares, or Fibonacci.
Always verify your pattern by checking whether it correctly predicts a term you already know.
:::

Unit Digit Patterns (Cyclicity)

Notes

You will never actually compute 7⁹⁵ in an exam hall — but you can find its unit digit in under 10 seconds once you understand cyclicity, and that skill alone is worth 2–3 marks per paper.

Definition: The unit digit of a number is its ones-place digit (the digit in the rightmost position). For any power aⁿ, the unit digit depends only on the unit digit of a and the value of n — never on the full value of a.

Definition: Cyclicity refers to the fact that as n increases, the unit digit of aⁿ repeats in a fixed cycle. The length of this cycle is called the cycle length (or period).

Why Unit Digit Patterns Work

When you compute successive powers of any number, the unit digit of the result depends only on the unit digit of the base and the unit digit of the previous result. For example:

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 ← cycle restarts!

The pattern (7, 9, 3, 1) repeats with period 4. You never need to compute the actual value.

The Five Cyclicity Cases

Case 1: Digits 0, 1, 5, 6 — Cycle Length 1

These digits are fixed point digits — raising them to any power never changes the unit digit.

Digit	Any power	Unit digit
0	0ⁿ	0
1	1ⁿ	1
5	5ⁿ	5
6	6ⁿ	6

Real-world example: 256¹⁰⁰ ends in 6. 5⁵⁰⁰ ends in 5. No calculation required.

Why: 5² = 25, 5³ = 125, 5⁴ = 625 — unit digit is always 5. This is because 5 × 5 = 25, always ending in 5 again. Similarly, 6 × 6 = 36, always ending in 6.

Case 2: Digit 4 — Cycle Length 2

4¹ = 4, 4² = 16, 4³ = 64, 4⁴ = 256 → cycle: (4, 6, 4, 6, ...)

Rule: odd power → unit digit 4; even power → unit digit 6.

Power (n)	n is odd	n is even
Unit digit of 4ⁿ	4	6

Case 3: Digit 9 — Cycle Length 2

9¹ = 9, 9² = 81, 9³ = 729, 9⁴ = 6561 → cycle: (9, 1, 9, 1, ...)

Rule: odd power → unit digit 9; even power → unit digit 1.

Case 4: Digits 2, 3, 7, 8 — Cycle Length 4

These are the most tested digits in exams:

Digit	Position 1	Position 2	Position 3	Position 4
2ⁿ	2	4	8	6
3ⁿ	3	9	7	1
7ⁿ	7	9	3	1
8ⁿ	8	4	2	6

Memory pattern for 2: 2 → 4 → 8 → 6 (even powers show 4, 6 alternating; odd powers show 2, 8 alternating).

Memory pattern for 7: Note that 7 and 3 have the same cycle elements (7,9,3,1 and 3,9,7,1) — just reversed starting points.

The Universal Method: Using Mod 4

For digits with cycle length 4 (2, 3, 7, 8), the position in the cycle is determined by n mod 4:

n mod 4	Position in 4-cycle
1	1st element of cycle
2	2nd element of cycle
3	3rd element of cycle
0 (i.e., divisible by 4)	4th element of cycle

Key rule: When n mod 4 = 0, use the 4th position (not the 0th — cycles are 1-indexed).

Step-by-Step Procedure

Identify the unit digit of the base (ignore all other digits).
Determine the cycle for that digit.
If cycle length is 1: done.
If cycle length is 2: check if exponent is odd or even.
If cycle length is 4: compute n mod 4; if result is 0, use position 4.
Read off the unit digit from that position.

Worked Example 1

Question: Find the unit digit of 2⁸³.

Solution:
Step 1: Unit digit of base = 2; cycle for 2 = (2, 4, 8, 6), length 4.
Step 2: 83 mod 4 = 3 (since 83 = 4×20 + 3).
Step 3: Position 3 in cycle (2, 4, 8, 6) → unit digit = 8.
Conclusion: Unit digit of 2⁸³ is 8.

Worked Example 2

Question: Find the unit digit of 3¹⁰⁰.

Solution:
Step 1: Cycle for 3 = (3, 9, 7, 1), length 4.
Step 2: 100 mod 4 = 0.
Step 3: n mod 4 = 0 → use position 4 → unit digit = 1.
Conclusion: Unit digit of 3¹⁰⁰ is 1.

Worked Example 3

Question: Find the unit digit of 9⁴⁷.

Solution:
Step 1: Cycle for 9 = (9, 1), length 2.
Step 2: 47 is odd → unit digit = 9.
Conclusion: Unit digit of 9⁴⁷ is 9.

Multi-Term Unit Digit Problems

Exams also ask for the unit digit of expressions like 2⁴³ × 3³² + 7¹¹.

Approach: Find the unit digit of each part separately, then apply the operation to just those unit digits.

Question: Unit digit of 2⁴³ × 3³²?

2⁴³: 43 mod 4 = 3 → position 3 in (2,4,8,6) → 8.
3³²: 32 mod 4 = 0 → position 4 in (3,9,7,1) → 1.
Unit digit of product = unit digit of 8 × 1 = 8.

:::compare Cyclicity Quick Reference

Digit	Cycle	Length	Method
0	0	1	Always 0
1	1	1	Always 1
5	5	1	Always 5
6	6	1	Always 6
4	4, 6	2	Odd→4, Even→6
9	9, 1	2	Odd→9, Even→1
2	2, 4, 8, 6	4	n mod 4
3	3, 9, 7, 1	4	n mod 4
7	7, 9, 3, 1	4	n mod 4
8	8, 4, 2, 6	4	n mod 4
:::

:::keypoints Key points

Unit digit depends only on the unit digit of the base — ignore all other digits.
Digits 0, 1, 5, 6 always produce the same unit digit regardless of power.
Digits 4 and 9 have cycles of length 2: parity of exponent determines the unit digit.
Digits 2, 3, 7, 8 have cycles of length 4: use n mod 4, mapping 0 → position 4.
For multi-term expressions, find each term's unit digit, then combine using the operation.
The cycles for 2 and 8 are reverses of each other: (2,4,8,6) vs (8,4,2,6).
The cycles for 3 and 7 are reverses of each other: (3,9,7,1) vs (7,9,3,1).
Memorise cycles as 4-digit numbers: 2486, 3971, 7931, 8426.
:::

:::memory
"Zero Is Sure, Five Stays Six" → 0,1,5,6 never change. For the cycle-4 group: "Two Four Eight Six" for digit 2 (then read 3,7,8 as rotations). Remember: 49 is 4×4's friend (4's cycle is 4,6) and 9's cycle is 9 then 1.
:::

:::recap

Cyclicity is the pattern of repeating unit digits in successive powers of a base digit.
Digits 0,1,5,6 have cycle length 1; 4 and 9 have cycle length 2; 2,3,7,8 have cycle length 4.
For cycle-4 digits, compute n mod 4 to find the cycle position (0 maps to position 4).
For cycle-2 digits (4 and 9), check only whether the exponent is odd or even.
To find the unit digit of a product or sum of powers, find each unit digit separately first.
Memorise the four 4-digit cycles: 2→2486, 3→3971, 7→7931, 8→8426.
:::

Worked Example: Unit Digit and Series

Worked example

The best way to lock in a concept is to watch it solve a real problem — these worked examples stitch together unit digit cyclicity and odd-number series into the exact question formats SSC and RRB examiners favour.

Problem Set A: Unit Digit of a Large Power

Worked Example 1

Question: Find the unit digit of 7⁹⁵.

Solution:

Step 1: Identify the base's unit digit. The base is 7 — unit digit is 7.

Step 2: Recall the cycle for 7. Powers of 7 produce unit digits: 7, 9, 3, 1, 7, 9, 3, 1, … The cycle length is 4.

Step 3: Find the position in the cycle. Compute 95 mod 4.
95 = 4 × 23 + 3, so 95 mod 4 = 3.

Step 4: Read position 3 from the cycle (7, 9, 3, 1). Unit digit = 3.

Conclusion: The unit digit of 7⁹⁵ is 3.

Why this works: The unit digit of 7ⁿ is entirely determined by where n falls in the 4-step repeating cycle. After every 4 steps the pattern resets. 95 = 92 (divisible by 4) + 3, so 7⁹⁵ has the same unit digit as 7³ = 343, which ends in 3.

Worked Example 2

Question: Find the unit digit of 8¹⁰².

Solution:

Step 1: Cycle for 8 is (8, 4, 2, 6), length 4.

Step 2: 102 mod 4 = 2 (since 102 = 4×25 + 2).

Step 3: Position 2 in cycle (8, 4, 2, 6) → unit digit = 4.

Conclusion: The unit digit of 8¹⁰² is 4.

Worked Example 3 (Combined Expression)

Question: Find the unit digit of 3⁸⁴ + 6⁷⁵.

Solution:

Step 1 (3⁸⁴): Cycle for 3 is (3, 9, 7, 1), length 4. 84 mod 4 = 0 → position 4 → unit digit of 3⁸⁴ = 1.

Step 2 (6⁷⁵): Digit 6 always has unit digit 6 regardless of power.

Step 3: Unit digit of sum = unit digit of (1 + 6) = 7.

Conclusion: The unit digit of 3⁸⁴ + 6⁷⁵ is 7.

Common misconception corrected: When adding two large powers, you cannot add the full numbers — but you CAN add their unit digits to get the unit digit of the sum (as long as you look only at the ones place of the result). This is valid because (…a) + (…b) always ends in the unit digit of a+b.

Problem Set B: Sum of an Odd Number Series

Worked Example 4

Question: Find the sum 1 + 3 + 5 + 7 + … + 99.

Solution:

Step 1: Confirm this is a sum of consecutive odd numbers starting at 1.

Step 2: Count the number of terms. The kth odd number is 2k − 1. Set 2k − 1 = 99:
2k = 100 → k = 50.
So there are 50 terms.

Step 3: Apply the formula: Sum of first n odd numbers = n².
Sum = 50² = 2500.

Conclusion: 1 + 3 + 5 + 7 + … + 99 = 2500.

Why n² works: Pair the terms from both ends: (1 + 99) = 100, (3 + 97) = 100, … There are 25 such pairs, total = 2500. This confirms n² = 50² = 2500.

Real-world Indian analogy: Imagine laying tiles in a square courtyard. Row 1 has 1 tile, row 2 has 3 tiles (the L-shaped border), row 3 adds 5 more, and so on. After n rows the total is always n² tiles — the sum of first n odd numbers literally builds a square.

Worked Example 5 (Odd numbers not starting from 1)

Question: Find the sum of all odd numbers from 21 to 51.

Solution:

Step 1: The series is 21, 23, 25, ..., 51. This is an AP with a = 21, d = 2.

Step 2: Number of terms: n = (51 − 21)/2 + 1 = 30/2 + 1 = 16.

Step 3: Sum = (n/2) × (first + last) = (16/2) × (21 + 51) = 8 × 72 = 576.

Conclusion: Sum = 576.

Alternative method: Sum of odd numbers up to 51 = 26² = 676 (there are 26 odd numbers from 1 to 51). Sum of odd numbers up to 19 = 10² = 100 (there are 10 odd numbers from 1 to 19). Difference = 676 − 100 = 576. ✓

Problem Set C: Combined Quick-Fire Questions

Worked Example 6

Question: What is the unit digit of (3 × 7)⁴⁸?

Solution:

Step 1: 3 × 7 = 21, unit digit = 1.

Step 2: Digit 1 raised to any power always gives unit digit 1.

Conclusion: Unit digit = 1.

Shortcut: You don't even need to compute 3 × 7 fully; just note that the unit digit of 21 is 1, and 1ⁿ = 1 always.

Worked Example 7

Question: The sum of the first 20 odd numbers is how much more than the sum of the first 15 odd numbers?

Solution:

Step 1: Sum of first 20 odd numbers = 20² = 400.
Step 2: Sum of first 15 odd numbers = 15² = 225.
Step 3: Difference = 400 − 225 = 175.

Conclusion: The sum of the first 20 odd numbers exceeds the sum of the first 15 odd numbers by 175.

:::compare Unit Digit Approach vs. Direct Computation

Task	Direct computation	Unit digit method
7⁹⁵	Compute 7⁹⁵ — impossibly large	95 mod 4 = 3 → cycle pos 3 → 3
3⁸⁴ + 6⁷⁵	Impossible without a calculator	1 + 6 = 7 (unit digit)
Sum 1+3+…+99	Add 50 terms manually	n=50, 50²=2500
:::

:::keypoints Key points

The unit digit of 7ⁿ is found by n mod 4; the cycle is (7, 9, 3, 1).
When n mod 4 = 0, always use position 4, not position 0.
For combined expressions (sum/product), find each term's unit digit, then combine.
Sum of the first n odd numbers = n²; to identify n, solve 2n−1 = last term.
For odd-number series not starting at 1, use the AP formula S = (n/2)(first + last).
Digit 6 always produces unit digit 6; digit 1 always produces unit digit 1.
The "square pairing" technique (matching first with last) independently confirms n².
Each formula is fast — the key is instant recall of the cycle or the formula.
:::

:::memory
"Seven Nines Three One" — the cycle for 7 as a spoken chant. For odd-number sums: "Count, Square, Done" — count how many odd numbers, square that count.
:::

:::recap

Unit digit of 7⁹⁵: cycle (7,9,3,1), 95 mod 4 = 3, answer = 3.
For digits 2, 3, 7, 8: compute n mod 4; map 0 to position 4 in the 4-step cycle.
Sum 1+3+5+…+(2n−1) = n²; find n by solving 2n−1 = last odd term.
For odd sums not starting at 1: use the AP sum formula (n/2)(first+last).
Combined unit digit problems: evaluate each component separately, then combine.
Digit 6 (and 1, 5, 0) never change their unit digit — instant answer for bases ending in these.
:::