Module 3 · SSC CGL Quantitative Aptitude

Number System

Divisibility, LCM-HCF, remainder, simplification.

Divisibility · LCM & HCF · Simplification

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Learning Objectives

Understand the fundamental concepts of Number System
Apply key formulas and techniques to solve problems
Practice with exam-level questions to build speed and accuracy

Introduction

The Number System is the bedrock of the Quantitative Aptitude section in SSC CGL. It forms the foundation for nearly every other topic — from percentages and ratios to algebra and data interpretation. Mastering the number system not only helps you solve direct questions but also improves your calculation speed and number sense across the board.

In SSC CGL Tier 1, you can expect 3–5 questions directly from number system concepts. In Tier 2, the questions are more integrated — combining divisibility with remainders, or LCM-HCF with fractions. A strong grasp of this topic is essential for scoring 45+ in the quant section.

SSC CGL Exam Weightage

Over the past 5 years, the topic-wise distribution for Number System in SSC CGL has been consistent:

Exam	Questions	Topics Frequently Asked
Tier 1 (2020–2025)	3–5	Divisibility, Unit Digit, LCM-HCF, Simplification
Tier 2 (Paper 1)	4–7	Remainder Theorem, Cyclicity, Surds & Indices, Number of Factors
Tier 2 (Advanced)	2–3	Euler's Theorem, Wilson's Theorem, Base Conversion

Key insight: Divisibility and unit digit questions appear in almost every SSC CGL exam. These are high-weightage, low-effort questions — if you know the rules, you can solve them in under 10 seconds.

1. Types of Numbers

Understanding the classification of numbers is the first step in mastering the number system. Each category has distinct properties that are frequently tested.

Natural Numbers

Natural numbers are the counting numbers starting from 1: {1, 2, 3, 4, ...}. They are denoted by N. Key properties: (a) They are closed under addition and multiplication. (b) They are not closed under subtraction or division. For example, 3 − 5 = −2 is not a natural number.

Whole Numbers

Whole numbers include all natural numbers along with zero: {0, 1, 2, 3, ...}. Denoted by W. The only difference from natural numbers is the inclusion of zero. Zero is neither positive nor negative but is an even integer.

Integers

Integers include all whole numbers and their negatives: {..., −3, −2, −1, 0, 1, 2, 3, ...}. Denoted by Z. Integers are closed under addition, subtraction, and multiplication, but not under division.

Rational & Irrational Numbers

Rational numbers (Q) can be expressed as p/q where p and q are integers and q ≠ 0. Examples: 2/3, −4/5, 7, 0.25. Every rational number has either terminating or non-terminating repeating decimal expansion. Example: 1/3 = 0.3333..., 1/2 = 0.5.

Irrational numbers cannot be expressed as p/q. Their decimal expansion is non-terminating and non-repeating. Examples: √2, √3, π, e. Important: The sum or product of a rational and an irrational number is always irrational. Example: 2 + √3 is irrational.

Prime, Composite & Co-prime Numbers

Prime numbers have exactly two distinct factors: 1 and the number itself. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Note: 2 is the only even prime number. 1 is neither prime nor composite.

To test if a number N is prime, check divisibility by all primes up to √N. For example, to check if 97 is prime, test divisibility by 2, 3, 5, 7 (since √97 ≈ 9.8). 97 is not divisible by any of these, so it is prime.

Composite numbers have more than two factors. Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16.

Co-prime numbers (or relatively prime) have HCF = 1. They need not be individually prime. Examples: (4, 9), (8, 15), (14, 25). Note: Two consecutive numbers are always co-prime. Example: 14 and 15 are co-prime.

Even & Odd Numbers

Even numbers are divisible by 2 (last digit 0, 2, 4, 6, 8). Odd numbers are not divisible by 2 (last digit 1, 3, 5, 7, 9). Key properties: Even + Even = Even; Odd + Odd = Even; Even + Odd = Odd; Even × Even = Even; Odd × Odd = Odd; Even × Odd = Even.

Pro Tip: Prime Number Identification

Memorise all primes up to 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. In SSC CGL, questions often hinge on recognising whether a number in the 50–100 range is prime. Knowing this list saves 15–20 seconds per question.

Example: Prime Identification

Which of the following is a prime number?

a) 87 b) 91 c) 97 d) 99

Solution: c) 97. 87 = 3 × 29; 91 = 7 × 13; 99 = 9 × 11. Only 97 has no factors other than 1 and itself.

2. Divisibility Rules

Divisibility rules are shortcuts to determine whether a number is divisible by another number without performing full division. These are among the most frequently tested concepts in SSC CGL Tier 1.

Divisibility by 2

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). Example: 1234 is divisible by 2 (last digit 4). 1235 is not.

Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3. Example: 12345 → 1+2+3+4+5 = 15 → divisible by 3. So 12345 is divisible by 3.

Divisibility by 4

A number is divisible by 4 if its last two digits are divisible by 4. Example: 73124 → last two digits 24 → divisible by 4. So 73124 is divisible by 4.

Divisibility by 5

A number is divisible by 5 if its last digit is 0 or 5. Example: 12345 (ends with 5) and 12340 (ends with 0) are divisible by 5.

Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 and 3. Example: 12342 is even and sum of digits = 12 (divisible by 3), so it is divisible by 6.

Divisibility by 7

There are two popular methods:
Method 1 (Chunking): Double the last digit and subtract it from the remaining number. Repeat until you get a small number. If that number is divisible by 7, the original is divisible by 7.
Example: 343 → 34 − (2×3) = 34 − 6 = 28 → divisible by 7. So 343 is divisible by 7.
Method 2 (Osculation): Multiply the last digit by 5 and add to the rest. 343 → 34 + (5×3) = 34 + 15 = 49 → divisible by 7.

Divisibility by 8

A number is divisible by 8 if its last three digits are divisible by 8. Example: 73124 → last three digits 124 → 124/8 = 15.5, not divisible. 73128 → 128/8 = 16, so divisible by 8.

Divisibility by 9

A number is divisible by 9 if the sum of its digits is divisible by 9. Example: 12345 → 1+2+3+4+5 = 15 → not divisible by 9. 123456 → sum = 21 → not divisible. 12345678 → sum = 36 → divisible by 9.

Divisibility by 10

A number is divisible by 10 if its last digit is 0. Example: 12340 is divisible by 10; 12345 is not.

Divisibility by 11

A number is divisible by 11 if the difference between the sum of digits at odd positions (from the right) and the sum of digits at even positions is 0 or a multiple of 11. Example: 13321 → Odd positions (from right): 1+3+1 = 5; Even positions: 3+2 = 5; Difference = 0 → divisible by 11.

Combined Divisibility

SSC CGL frequently asks combined divisibility questions where a number must be divisible by two or more numbers simultaneously. For example, to be divisible by both 8 and 9, a number must satisfy both divisibility conditions. To be divisible by 12, it must be divisible by both 3 and 4.

Pro Tip: Divisibility by 13

Multiply the last digit by 4 and add to the remaining number. Repeat. If the result is divisible by 13, so is the original. Example: 169 → 16 + (4×9) = 16 + 36 = 52 → divisible by 13. This trick is not commonly known but appears in advanced SSC CGL Tier 2 questions.

Example: Combined Divisibility

Find the smallest number that is divisible by both 8 and 9.

a) 36 b) 48 c) 72 d) 96

Solution: c) 72. For a number to be divisible by both 8 and 9, it must be divisible by LCM(8,9) = 72. Checking: 72/8 = 9, 72/9 = 8.

Example: Divisibility by 7

Is 3913 divisible by 7?

a) Yes b) No

Solution: a) Yes. Using the rule: 3913 → 391 − (2×3) = 391 − 6 = 385 → 38 − (2×5) = 38 − 10 = 28 → divisible by 7. Hence 3913 is divisible by 7.

3. LCM & HCF

LCM (Least Common Multiple) and HCF (Highest Common Factor, also called GCD) are core concepts that appear in 1–2 questions in every SSC CGL Tier 1 exam. They also form the basis for time-work, time-speed-distance, and fraction problems.

Finding LCM

LCM of two or more numbers is the smallest positive number that is divisible by all of them.

Prime Factorization Method: Write each number as a product of primes. LCM is the product of the highest powers of all primes present.
Example: LCM of 24, 36, 48
24 = 2³ × 3; 36 = 2² × 3²; 48 = 2⁴ × 3
LCM = 2⁴ × 3² = 16 × 9 = 144

Division Method: Write numbers in a row. Divide by common prime factors. Continue until no common factor remains. Multiply all divisors and the final row.
Example: LCM(24, 36, 48):
2 | 24, 36, 48
2 | 12, 18, 24
2 | 6, 9, 12
3 | 3, 9, 6
| 1, 3, 2
LCM = 2 × 2 × 2 × 3 × 1 × 3 × 2 = 144

Finding HCF

HCF of two or more numbers is the largest number that divides all of them.

Prime Factorization Method: Take the product of the smallest powers of common primes.
Example: HCF of 24, 36, 48
24 = 2³ × 3; 36 = 2² × 3²; 48 = 2⁴ × 3
Common primes: 2 (min power 2² = 4), 3 (min power 3¹ = 3)
HCF = 4 × 3 = 12

Division Method: For two numbers, divide the larger by the smaller. Then divide the divisor by the remainder. Repeat until remainder is 0. The last divisor is the HCF.
Example: HCF of 84 and 144
144 ÷ 84 → remainder 60
84 ÷ 60 → remainder 24
60 ÷ 24 → remainder 12
24 ÷ 12 → remainder 0
HCF = 12

LCM & HCF of Fractions

LCM of Fractions = LCM of Numerators / HCF of Denominators
HCF of Fractions = HCF of Numerators / LCM of Denominators
Example: LCM of 2/3, 4/5, 6/7 = LCM(2,4,6) / HCF(3,5,7) = 12 / 1 = 12
Example: HCF of 2/3, 4/5, 6/7 = HCF(2,4,6) / LCM(3,5,7) = 2 / 105 = 2/105

Important Formulas

1. For two numbers a and b: a × b = LCM(a,b) × HCF(a,b)
2. HCF of two numbers is always ≤ the smaller number
3. LCM of two numbers is always ≥ the larger number
4. For co-prime numbers, HCF = 1, LCM = product of the numbers
5. LCM of consecutive numbers = their product; HCF = 1

Relation between LCM and HCF

The product of LCM and HCF of two numbers equals the product of the numbers. This is the most important formula in this chapter. It allows you to find one if you know the other. Example: If LCM = 360 and one number = 60, then the other number = (LCM × HCF) / 60. If HCF = 12, then other = (360 × 12) / 60 = 72.

Pro Tip: HCF using Difference

For any two numbers a and b, HCF(a,b) = HCF(a−b, b) for a > b. Example: HCF(144, 84) = HCF(60, 84) = HCF(60, 24) = HCF(36, 24) = HCF(12, 24) = 12. This is much faster than prime factorization for large numbers.

Example: LCM & HCF Formula

The LCM of two numbers is 240 and their HCF is 8. If one number is 40, find the other.

a) 48 b) 56 c) 64 d) 72

Solution: a) 48. Product = LCM × HCF = 240 × 8 = 1920. Other number = 1920 ÷ 40 = 48.

Example: LCM of Fractions

Find the LCM of 3/4, 5/6, and 7/8.

Solution: LCM of fractions = LCM(3,5,7) / HCF(4,6,8) = 105 / 2 = 52.5.

Example: HCF of Three Numbers

Find the HCF of 96, 144, and 240.

Solution: Prime factors: 96 = 2⁵ × 3; 144 = 2⁴ × 3²; 240 = 2⁴ × 3 × 5. Common: 2⁴ × 3¹ = 16 × 3 = 48. HCF = 48.

4. Unit Digit Cyclicity

Unit digit questions are a staple of SSC CGL. The key insight: when calculating the unit digit of a power, only the unit digit of the base matters. All other digits can be ignored.

Cyclicity Table

Every digit from 0–9 has a repeating cycle for its powers:

Digit	Cycle	Cycle Length
0	0	1
1	1	1
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
4	4, 6	2
5	5	1
6	6	1
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4
9	9, 1	2

Finding Unit Digit of Large Powers

Step 1: Identify the unit digit of the base.
Step 2: Find the cyclicity length of that digit (from table above).
Step 3: Divide the exponent by the cyclicity length.
Step 4: If remainder = r, the unit digit = r-th term in the cycle. If remainder = 0, use the last term of the cycle.

Example: Unit digit of 7²⁵: Cyclicity of 7 = 4. 25 ÷ 4 = 6 remainder 1. So unit digit = 7¹ = 7.

Unit Digit in Factorials

The unit digit of any factorial n! for n ≥ 5 is always 0, because 5! = 120 includes factors 2 and 5, giving a trailing zero. For sums of factorials, only compute the unit digits of 1! through 4!, then add them.

Pro Tip: Unit Digit of Sums

When finding the unit digit of a sum (e.g., a^m + b^n), find the unit digit of each term separately using cyclicity, then add the unit digits. Example: 7²⁵ + 3¹⁵ → unit digit of 7²⁵ = 7, unit digit of 3¹⁵ = 7 (since 15÷4=3 rem 3 → 3³ cycle = 7). Sum = 7+7 = 14 → unit digit = 4.

Example: Unit Digit of Large Power

Find the unit digit of 2⁵⁰.

a) 2 b) 4 c) 6 d) 8

Solution: b) 4. Cyclicity of 2: 2,4,8,6 (length 4). 50 ÷ 4 = 12 remainder 2. The 2nd term in cycle = 4.

Example: Unit Digit of Sum of Factorials

Find the unit digit of 1! + 2! + 3! + 4! + 5! + ... + 50!.

Solution: For n ≥ 5, n! ends with 0. So only compute 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 → unit digit = 3.

Example: Sum of Powers

Find the unit digit of 3⁴² + 7²³.

a) 2 b) 4 c) 6 d) 8

Solution: b) 4. Cyclicity of 3: 3,9,7,1 (length 4). 42 ÷ 4 = 10 rem 2 → unit digit = 9. Cyclicity of 7: 7,9,3,1 (length 4). 23 ÷ 4 = 5 rem 3 → unit digit = 3. Sum unit digit = 9 + 3 = 12 → unit digit = 2.

5. Remainder Theorem

Remainder questions appear in SSC CGL Tier 2 regularly. Understanding remainders modulo a divisor is crucial for advanced number theory questions.

Basic Remainder Theorem

When a number N is divided by d, it can be written as N = d × q + r, where 0 ≤ r < d. The remainder r is unique. Key property: (a + b) mod d = [(a mod d) + (b mod d)] mod d. Similarly, (a × b) mod d = [(a mod d) × (b mod d)] mod d.

Negative Remainder Concept

A negative remainder is simply the divisor minus the positive remainder. For example, when dividing 23 by 7, the remainder is 2. The negative remainder is 2 − 7 = −5. This is useful because for large exponents, negative remainders can simplify calculations. Example: 2⁴⁵ mod 7: Since 2³ ≡ 1 (mod 7), 2⁴⁵ = (2³)¹⁵ ≡ 1¹⁵ ≡ 1 (mod 7). Remainder = 1.

Euler's Theorem

Euler's theorem states that for any integer a coprime to n, a^φ(n) ≡ 1 (mod n), where φ(n) is Euler's totient function (number of integers from 1 to n that are coprime to n). For a prime p, φ(p) = p − 1. Example: φ(7) = 6, so a⁶ ≡ 1 (mod 7) for any a not divisible by 7.

Fermat's Little Theorem

A special case of Euler's theorem for prime moduli: if p is prime and a is not divisible by p, then a^(p−1) ≡ 1 (mod p). Example: 2⁶ ≡ 1 (mod 7), which checks because 2⁶ = 64 and 64 mod 7 = 1.

Wilson's Theorem

For a prime number p, (p−1)! ≡ −1 (mod p). Equivalently, (p−1)! + 1 is divisible by p. Example: 4! = 24, and 24 + 1 = 25 is divisible by 5. So 4! ≡ −1 (mod 5) or 4! ≡ 4 (mod 5).

Pro Tip: Finding Remainders of Large Powers

When the divisor is small (like 5, 7, 8), find the cyclicity of the base modulo the divisor. For example, 2 mod 5 cycles as 2,4,3,1 (length 4). So 2^100 mod 5 = 1 (since 100 is divisible by 4). This is faster than applying Euler's theorem for small divisors.

Example: Basic Remainder

Find the remainder when 3⁵⁰ is divided by 5.

Solution: 3 mod 5 cycles: 3¹≡3, 3²≡4, 3³≡2, 3⁴≡1 (length 4). 50 ÷ 4 = 12 rem 2. So 3⁵⁰ ≡ 3² ≡ 4 (mod 5). Remainder = 4.

Example: Negative Remainder

Find the remainder when 2⁴⁵ is divided by 7.

Solution: 2³ = 8 ≡ 1 (mod 7). So 2⁴⁵ = (2³)¹⁵ ≡ 1¹⁵ = 1 (mod 7). Remainder = 1.

Example: Euler's Theorem

Find the remainder when 5¹⁰⁰ is divided by 12.

Solution: gcd(5,12) = 1. φ(12) = 12 × (1−1/2) × (1−1/3) = 12 × 1/2 × 2/3 = 4. By Euler's theorem, 5⁴ ≡ 1 (mod 12). 100 = 4 × 25, so 5¹⁰⁰ = (5⁴)²⁵ ≡ 1²⁵ = 1 (mod 12). Remainder = 1.

6. Simplification & Approximation

Simplification questions test your ability to apply the correct order of operations and approximate values quickly. These are among the easiest marks in SSC CGL if you avoid common mistakes.

BODMAS Rule

The order of operations is: Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction. SSC CGL questions often include nested brackets. Always solve from the innermost bracket outward.

Example: 12 + [8 − (4 ÷ 2) × 3] = 12 + [8 − 2 × 3] = 12 + [8 − 6] = 12 + 2 = 14.

Approximation Techniques

For questions where the options are far apart, round numbers to the nearest convenient value. Example: 41.8 × 19.2 ≈ 42 × 19 = 798. The actual value is 802.56, and if options are 750, 800, 850, 900, you can confidently select 800.

Fraction to Percentage Conversion

Memorise this table for quick conversions:

Fraction	Percentage
1/2	50%
1/3	33.33%
1/4	25%
1/5	20%
1/6	16.67%
1/7	14.28%
1/8	12.5%
1/9	11.11%
1/10	10%
1/11	9.09%
1/12	8.33%

Pro Tip: Simplification Shortcut

In SSC CGL, simplification questions often have a common factor. Factorise and cancel before calculating. Example: (48² − 36²) / 12 = (48−36)(48+36) / 12 = (12 × 84) / 12 = 84. This is much faster than computing squares.

Example: BODMAS

Simplify: 15 − [3 + 4 × (8 ÷ 2) − 7]

Solution: Innermost bracket: 8 ÷ 2 = 4. Then: 3 + 4 × 4 − 7 = 3 + 16 − 7 = 12. Finally: 15 − 12 = 3.

Example: Approximation

Approximate: 24.8% of 489.5 + 35.2% of 612.3

Solution: 24.8% ≈ 1/4, 35.2% ≈ 1/3. Approx = 490/4 + 612/3 = 122.5 + 204 = 326.5. Actual is close to 327.

7. Surds & Indices

Surds and Indices questions test your ability to manipulate expressions involving roots and powers. These appear in both Tier 1 and Tier 2.

Laws of Indices

1. a^m × a^n = a^(m+n)
2. a^m ÷ a^n = a^(m−n)
3. (a^m)^n = a^(m×n)
4. a^0 = 1 (for a ≠ 0)
5. a^(−n) = 1 / a^n
6. a^(m/n) = (a^m)^(1/n) = nth root of a^m
7. (a × b)^n = a^n × b^n
8. (a/b)^n = a^n / b^n

Laws of Surds

1. √(a × b) = √a × √b (for a, b ≥ 0)
2. √(a / b) = √a / √b (for a ≥ 0, b > 0)
3. (√a)^2 = a
4. a√c + b√c = (a+b)√c (only like surds can be added)
5. Rationalization: 1/(√a + √b) = (√a − √b)/(a − b)

Rationalization

Rationalizing the denominator is a common technique. Multiply the numerator and denominator by the conjugate of the denominator. Example: 1/(√3 + √2) = (√3 − √2)/(3 − 2) = √3 − √2.

Comparison of Surds

To compare surds like √5, ∛10, ∜17, raise all to the power of LCM of their indices. LCM of 2,3,4 = 12. √5 = 5^(6/12) = 15625^(1/12); ∛10 = 10^(4/12) = 10000^(1/12); ∜17 = 17^(3/12) = 4913^(1/12). Since 15625 > 10000 > 4913, we have √5 > ∛10 > ∜17.

Pro Tip: Rationalization Trick

When the denominator is of the form √a + √b + √c, group two terms and rationalize step by step. Example: 1/(√2 + √3 + √5) → first rationalize 1/((√2+√3)+√5). This is a common Tier 2 pattern.

Example: Laws of Indices

Simplify: (2⁵ × 2⁸) / 2¹⁰

Solution: = 2^(5+8−10) = 2³ = 8.

Example: Rationalization

Rationalize: 1 / (√7 − √5)

Solution: Multiply numerator and denominator by (√7 + √5): = (√7 + √5) / (7 − 5) = (√7 + √5) / 2.

Example: Surd Comparison

Which is larger: ∛4 or √2?

Solution: LCM of indices 2 and 3 = 6. ∛4 = 4^(2/6) = 16^(1/6). √2 = 2^(3/6) = 8^(1/6). Since 16 > 8, ∛4 > √2.

8. Square Roots & Cube Roots

Square root and cube root questions test your ability to find roots of perfect squares/cubes and to estimate values for non-perfect ones.

Square Root Techniques

For a perfect square, pair digits from the right. The number of pairs gives the number of digits in the square root. Use the digit-by-digit method or prime factorization. Example: √1764 → prime factors: 1764 = 2² × 3² × 7² = (2×3×7)² = 42². So √1764 = 42.

Cube Root Techniques

For perfect cubes, group digits in threes from the right. The last digit of the cube root depends on the last digit of the cube: 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9, 0→0. Example: ∛4913 → last digit 3 → cube root ends with 7. 4913 lies between 8³=512 and 27³=19683, and 17³=4913. So ∛4913 = 17.

Approximation of Square Roots

For non-perfect squares, use the formula: √N ≈ (N + P²) / (2P), where P² is the nearest perfect square. Example: √40 → nearest perfect square = 36 (6²). √40 ≈ (40 + 36) / (2×6) = 76/12 ≈ 6.33. Actual √40 = 6.3249.

Pro Tip: Square Root of Large Perfect Squares

If a number ends with 25, it is the square of a number ending with 5. Example: 4225 → last two digits 25 → ends with 5. Remove 25, take √42 ≈ 6. So √4225 = 65 (since 6×7 = 42, prefix 6, suffix 5). This trick works for any (a5)² pattern.

Example: Cube Root

Find ∛12167.

Solution: Last digit 7 → cube root ends with 3. 12167 lies between 20³=8000 and 30³=27000. 23³ = 12167. Answer = 23.

Example: Square Root Approximation

Approximate √85 to one decimal place.

Solution: Nearest perfect square = 81 (9²). √85 ≈ (85 + 81) / (2×9) = 166/18 ≈ 9.22. Actual √85 = 9.2195.

9. Number of Factors

Finding the number of factors, sum of factors, and product of factors is a recurring theme in SSC CGL Tier 2.

Number of Factors

If N = a^p × b^q × c^r (where a, b, c are prime factors), then total number of factors = (p+1)(q+1)(r+1).
Example: N = 2³ × 3² × 5 = 360. Number of factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

Sum of Factors

Sum of factors = [(a^(p+1) − 1) / (a − 1)] × [(b^(q+1) − 1) / (b − 1)] × [(c^(r+1) − 1) / (c − 1)]
Example: For N = 2³ × 3² = 8 × 9 = 72, sum of factors = [(2⁴−1)/(2−1)] × [(3³−1)/(3−1)] = (16−1)/1 × (27−1)/2 = 15 × 13 = 195.

Number of Even/Odd Factors

Total even factors = Total factors − Odd factors. Odd factors come only from odd prime factors. Example: 360 = 2³ × 3² × 5. Odd factors from 3² × 5: (2+1)(1+1) = 6. Even factors = 24 − 6 = 18.

Example: Number of Factors

Find the total number of factors of 144.

a) 12 b) 15 c) 18 d) 20

Solution: b) 15. 144 = 2⁴ × 3². Number of factors = (4+1)(2+1) = 5 × 3 = 15.

Example: Even Factors

Find the number of even factors of 180.

Solution: 180 = 2² × 3² × 5. Total factors = (2+1)(2+1)(1+1) = 18. Odd factors (from 3² × 5) = (2+1)(1+1) = 6. Even factors = 18 − 6 = 12.

Shortcut Techniques

Pro Tip: Divisibility by 7

Double the last digit and subtract from the remaining number. If result is divisible by 7, the original is too. Example: 343 → 34 − (2×3) = 34 − 6 = 28 → divisible by 7. This can solve in 5 seconds vs 30 seconds with division.

Pro Tip: Divisibility by 13

Multiply the last digit by 4 and add to the remaining number. Repeat. Example: 169 → 16 + (4×9) = 16 + 36 = 52 → divisible by 13. This works for any number and is faster than trial division.

Pro Tip: Unit Digit of Powers Ending in 1, 5, 6

Any number ending in 1, 5, or 6 raised to any positive power has the same unit digit. Example: 51²³⁴ ends with 1, 25⁶⁷ ends with 5, 46⁸⁹ ends with 6. No calculation needed!

Pro Tip: HCF of Three Numbers by Difference

HCF(a, b, c) = HCF(HCF(a,b), c). But a faster method: HCF(a,b,c) = HCF(a−b, b−c, c). This is especially useful when numbers are close together. Example: HCF(96, 108, 120) → 108−96=12, 120−108=12, HCF(12,12)=12.

Pro Tip: Square of Numbers Ending in 5

For any number ending in 5 (like 35, 65, 115): Multiply the tens digit by (tens digit + 1), then append 25. Example: 35² → 3×4 = 12, append 25 = 1225. 65² → 6×7 = 42, append 25 = 4225. This works for any (a5)².

Pro Tip: Remainder When Divisor is 9

The remainder when a number is divided by 9 equals the digital root (sum of digits reduced to a single digit) unless the digital root is 0, in which case remainder is 9. Example: 12345 → sum = 15 → 1+5 = 6, so 12345 mod 9 = 6. This is much faster than actual division.

Pro Tip: Comparing Fractions

To compare fractions like 3/7 and 4/9, use cross-multiplication: 3×9 = 27, 4×7 = 28. Since 27 < 28, 3/7 < 4/9. This is faster than converting to decimals.

Pro Tip: Last Two Digits of Squares

The last two digits of a square depend only on the last two digits of the base. For example, all numbers ending with 24 have squares ending with 76 (24²=576, 124²=15376). Memorise common patterns: 0→00, 1→01, 2→04, 3→09, 4→16, 5→25, 6→36, 7→49, 8→64, 9→81.

Pro Tip: Digital Root for Checking Calculations

Use digital root (sum of digits reduced to a single digit) to verify multiplication or addition. Example: 45 × 67 = 3015. Digital root of 45 = 9, of 67 = 4 (6+7=13→1+3=4). 9×4=36→3+6=9. Digital root of 3015 = 3+0+1+5=9. They match, so the answer is likely correct.

Pro Tip: Finding Remainder Using Base Splitting

For large exponents like 2⁶⁰ mod 7, split the base: 2³ ≡ 1 (mod 7). Then 2⁶⁰ = (2³)²⁰ ≡ 1²⁰ = 1. This base-splitting trick works whenever the base modulo the divisor has a small cyclicity.

Pro Tip: Multiplying by 11

To multiply a two-digit number by 11, add the two digits and place the sum between them. Example: 34 × 11 → 3+4=7 → 374. If the sum exceeds 9, carry the 1: 76 × 11 → 7+6=13 → 836.

Pro Tip: Squaring Numbers Near 100

For numbers near 100 (like 97, 104): 97² = (97−3)² = 100−2(3)+9/100 format = 9409. More systematically: 97 = 100−3, so 97² = (100−3)² = 10000 − 600 + 9 = 9409. For 104: (100+4)² = 10000 + 800 + 16 = 10816.

Solved Examples (15 Examples)

Example 1: Divisibility by 11

Which of the following numbers is divisible by 11?

a) 12345 b) 12122 c) 13321 d) 14443

Solution: c) 13321. Sum of digits at odd places = 1+3+1 = 5. Sum at even places = 3+2 = 5. Difference = 0, divisible by 11.

Example 2: Unit Digit

Find the unit digit of 7²⁵.

a) 3 b) 5 c) 7 d) 9

Solution: c) 7. Cyclicity of 7 = 4. 25 ÷ 4 = 6 rem 1 → unit digit = 7¹ = 7.

Example 3: LCM of Three Numbers

Find the LCM of 24, 36, and 48.

a) 124 b) 144 c) 164 d) 188

Solution: b) 144. 24=2³×3, 36=2²×3², 48=2⁴×3. LCM = 2⁴×3² = 16×9 = 144.

Example 4: HCF using Formula

The LCM of two numbers is 360 and their HCF is 12. If one number is 60, find the other.

a) 72 b) 84 c) 96 d) 108

Solution: a) 72. Product = LCM × HCF = 360 × 12 = 4320. Other = 4320 ÷ 60 = 72.

Example 5: Sum of Unit Digits

What is the unit digit of (1234)¹⁰² + (5678)⁵⁸?

a) 0 b) 2 c) 4 d) 6

Solution: a) 0. 4¹⁰²: cyclicity of 4 is 2 (4,6). Even power → 6. 8⁵⁸: cyclicity of 8 is 4 (8,4,2,6). 58 ÷ 4 = 14 rem 2 → 4. Sum = 6+4 = 10 → unit digit 0.

Example 6: Remainder using Cyclicity

Find the remainder when 3⁵¹ is divided by 5.

Solution: 3 mod 5 cycles: 3,4,2,1 (length 4). 51 ÷ 4 = 12 rem 3 → 3rd term = 2. Remainder = 2.

Example 7: Divisibility by 8 and 9 Combined

If a number 774958A96B is divisible by 8 and 9, find (A + B).

Solution: For 8, last three digits 96B must be divisible by 8 → B = 0 or 8. Sum = 55+A+B. For 9, sum divisible by 9. If B=0 → 55+A → A=8 → A+B=8. If B=8 → 63+A → A=0 or 9 → A+B=8 or 17. The most consistent answer from the given options is 9.

Example 8: HCF of Fractions

Find the HCF of 3/4, 5/6, and 7/8.

Solution: HCF of fractions = HCF(3,5,7) / LCM(4,6,8) = 1 / 24.

Example 9: BODMAS Simplification

Simplify: 36 ÷ 4 × 3 + 5 − 2

Solution: By BODMAS, division first: 36 ÷ 4 = 9. Then multiplication: 9 × 3 = 27. Then addition: 27 + 5 = 32. Then subtraction: 32 − 2 = 30.

Example 10: Square Root of Large Number

Find √285156.

Solution: Pair from right: 28 | 51 | 56. √28 ≈ 5 (5²=25). Remainder 3. Bring down 51 → 351. Double quotient 5→10. 10_ × _ ≤ 351 → 103×3=309. Remainder 42. Bring down 56 → 4256. Double quotient 53→106. 106_ × _ ≤ 4256 → 1064×4=4256. Quotient = 534. So √285156 = 534.

Example 11: Number of Factors

Find the total number of factors of 300.

Solution: 300 = 2² × 3 × 5². Number of factors = (2+1)(1+1)(2+1) = 3×2×3 = 18.

Example 12: Surd Comparison

Which is largest: √6, ∛8, ∜5?

Solution: LCM of indices 2,3,4 = 12. √6 = 6^(6/12) = 46656^(1/12). ∛8 = 8^(4/12) = 4096^(1/12). ∜5 = 5^(3/12) = 125^(1/12). √6 is the largest.

Example 13: Approximation

Approximate: 19.8 × 31.2 ÷ 7.9

Solution: Round: 20 × 31 ÷ 8 = 620 ÷ 8 = 77.5. Actual = 78.2. Choose the option closest to 78.

Example 14: Factorial Unit Digit

Find the unit digit of 1! + 2! + 3! + 4! + ... + 25!.

Solution: n! for n ≥ 5 ends with 0. So sum = 1 + 2 + 6 + 24 = 33 → unit digit = 3.

Example 15: Euler's Theorem Application

Find the remainder when 3⁵⁰ is divided by 11.

Solution: φ(11) = 10. By Euler's theorem, 3¹⁰ ≡ 1 (mod 11). 50 = 10×5. So 3⁵⁰ = (3¹⁰)⁵ ≡ 1⁵ = 1 (mod 11). Remainder = 1.

Advanced Problems (Tier 2 Level)

Advanced 1: Remainder with Euler's Theorem

Find the remainder when 7²⁰²⁴ is divided by 12.

Solution: gcd(7,12) = 1. φ(12) = 4. 7⁴ ≡ 1 (mod 12). 2024 ÷ 4 = 506 exactly. So 7²⁰²⁴ = (7⁴)⁵⁰⁶ ≡ 1 (mod 12). Remainder = 1.

Advanced 2: Divisibility with Variables

If the number 5A6B2 is divisible by 9, find the minimum value of A + B.

Solution: Sum = 5+A+6+B+2 = 13+A+B. For divisibility by 9, 13+A+B must be 18 or 27. Minimum: 13+A+B = 18 → A+B = 5.

Advanced 3: Product of LCM and HCF

The LCM of two numbers is 60 times their HCF. If the sum of the numbers is 126 and their HCF is 6, find the numbers.

Solution: Let numbers be 6a and 6b (a,b co-prime). LCM = 6ab = 60×6 = 360 → ab = 60. Also 6a + 6b = 126 → a+b = 21. Solving: a=5, b=16 or a=16, b=5. Numbers are 30 and 96.

Advanced 4: Unit Digit of Multi-term Expression

Find the unit digit of (2²⁵ + 3³⁰ + 4³⁵) × 5⁴⁰.

Solution: 5⁴⁰ ends with 5. Any number multiplied by 5 gives 0 or 5. 2²⁵ → 2²⁵ cycle length 4: 25÷4=6rem1 → 2. 3³⁰ → 30÷4=7rem2 → 9. 4³⁵ → odd power → 4. Sum = 2+9+4=15 → unit digit 5. 5 × 5 = 25 → unit digit 5.

Advanced 5: Number of Zeroes in Factorial

Find the number of trailing zeroes in 100!.

Solution: Count factors of 5 in 100!: ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24. So 100! ends with 24 zeroes.

Advanced 6: Sum of Factors

Find the sum of all factors of 72.

Solution: 72 = 2³ × 3². Sum = [(2⁴−1)/(2−1)] × [(3³−1)/(3−1)] = (16−1)/1 × (27−1)/2 = 15 × 13 = 195.

Advanced 7: Wilson's Theorem

Find the remainder when 52! is divided by 53.

Solution: By Wilson's theorem, (p−1)! ≡ −1 (mod p) for prime p. Here p=53 is prime. So 52! ≡ −1 ≡ 52 (mod 53). Remainder = 52.

Common Mistakes & How to Avoid

Confusing Divisibility of 7 and 11: For 7, double and subtract. For 11, alternate sum difference. Many students mix these up. Practice both rules side by side until they become automatic.
Forgetting BODMAS Priority: Division and multiplication have equal priority (work left to right). Addition and subtraction also have equal priority (work left to right). Common error: 36 ÷ 4 × 3 = 36 ÷ 12 = 3 is WRONG. Correct: 36 ÷ 4 × 3 = 9 × 3 = 27.
Unit Digit of Powers with Exponent 0: Any non-zero number raised to 0 is 1, so the unit digit is always 1. But 0⁰ is undefined. Remember this edge case.
HCF vs LCM of Fractions: Students often invert the formula. Remember: LCM of fractions = LCM(num) / HCF(den); HCF of fractions = HCF(num) / LCM(den). The denominator operation is swapped.
Counting Factors of Zero: When asked for the number of trailing zeroes in n!, only count factors of 5 (since factors of 2 are always more abundant). Do not count both 2 and 5.
Negative Remainder Confusion: If remainder is negative (e.g., −2), the actual remainder is the divisor minus the absolute value (e.g., divisor − 2). Example: 5 mod 7 = −2 → actual remainder = 7 − 2 = 5.
Misapplying Euler's Theorem: Euler's theorem requires the base and modulus to be co-prime. If they are not co-prime, Euler's theorem does not apply directly.
Unit Digit of 0 or 1 Powers: 0 raised to any positive power is 0. 1 raised to any power is 1. But 0⁰ is undefined — SSC CGL rarely tests this, but be aware.
Using Division Method for HCF Incorrectly: The division method for HCF works only for two numbers. For three numbers, first find HCF of any two, then find HCF of that result with the third.
Not Simplifying Before Comparing Surds: When comparing surds with different indices, always raise to the LCM of the indices first. Comparing raw values leads to errors.
Forgetting 1 is Not Prime: Many students mistakenly treat 1 as a prime number. 1 has only one factor (itself) and is neither prime nor composite.
Confusing LCM with HCF in Word Problems: When a question asks for the greatest number that divides two numbers, it is HCF. When it asks for the smallest number divisible by two numbers, it is LCM. Read carefully.
Assuming Numbers Ending with 5 are Always Divisible by 5 Only: A number ending with 5 is divisible by 5 but may also be divisible by other numbers (e.g., 25 is divisible by 25, 125 is divisible by 125).
Unit Digit of Negative Bases: For negative bases with even exponents, the negative sign disappears. Example: (−3)⁴² has the same unit digit as 3⁴² = 9.
Not Considering Zero as a Possible Digit: In problems like "Find A+B if a number is divisible by 9", do not forget that A or B could be 0. Zero is a valid digit.
Mishandling Remainders in Combined Conditions: When a number leaves remainder r when divided by multiple divisors, use LCM of divisors + r only if the remainder is the same for all divisors.
Treating All Irrational Numbers as Surds: Not all irrational numbers are surds. A surd specifically refers to an irrational root (like √2, √3). Numbers like π or e are irrational but not surds.

Pro Tips from Toppers

Topper Tip 1: Time Management

In SSC CGL Tier 1, you have roughly 50 seconds per question. Divisibility, unit digit, and LCM-HCF questions should take 10–20 seconds max. If a question takes more than 30 seconds, flag it and move on. Come back only if you have time at the end.

Topper Tip 2: Question Selection

Not all questions are worth the same time. Prioritise: Divisibility rules (easiest) → Unit digit → LCM-HCF → Simplification → Remainder theorem → Surds & Indices. Skip questions with complex factorial expressions unless you are very comfortable.

Topper Tip 3: Rough Sheet Strategy

Write down the cyclicity table and common divisibility rules on your rough sheet as soon as the exam begins. This saves precious seconds from mental recall and prevents mistakes under pressure.

Topper Tip 4: Eliminate Before Solving

For MCQs, eliminate obviously wrong options first. Example: If the question asks for a prime number, eliminate all even numbers and all numbers divisible by 3 or 5 immediately. This narrows down to 1–2 options.

Topper Tip 5: Practice Mental Calculation

Practice squares up to 30, cubes up to 15, and fraction-to-percentage conversions daily. Students who have these memorised save 5–8 minutes in the quant section, which they can use on harder problems.

Topper Tip 6: Previous Year Paper Analysis

Analyse the last 10 SSC CGL papers. You will find that certain patterns repeat — for example, divisibility by 8&9 combined appears every 2–3 exams. Focus on these high-frequency patterns.

Topper Tip 7: Avoid Over-Complication

Most SSC CGL number system questions can be solved with basic rules — you do not need Euler's theorem or Wilson's theorem for Tier 1. Save advanced theorems for Tier 2 preparation. In Tier 1, stick to cyclicity and basic divisibility.

Topper Tip 8: Speed vs Accuracy Trade-off

Aim for 100% accuracy on the first 15 quant questions (which include number system). It is better to solve 20 questions with 95% accuracy than 25 questions with 80% accuracy, because SSC CGL has no negative marking for unanswered questions.

Topper Tip 9: Formula Sheet Revision

Create a one-page formula sheet for number system and revise it every morning 10 minutes before your practice session. Repetition is the key to instant recall in exam conditions.

Topper Tip 10: Mock Test Strategy

In mock tests, always attempt number system questions first since they are quick and high-confidence. A good score in the first 5 minutes builds momentum for the rest of the quant section.

Summary of Key Formulas

Concept	Formula
LCM of two numbers	LCM(a,b) = (a × b) / HCF(a,b)
HCF of two numbers	HCF(a,b) = (a × b) / LCM(a,b)
Product relation	a × b = LCM(a,b) × HCF(a,b)
LCM of fractions	LCM(Num) / HCF(Den)
HCF of fractions	HCF(Num) / LCM(Den)
Number of factors (N = aᵖ × bᵟ × cʳ)	(p+1)(q+1)(r+1)
Sum of factors	[(aᵖ⁺¹−1)/(a−1)] × [(bᵟ⁺¹−1)/(b−1)] × [(cʳ⁺¹−1)/(c−1)]
Trailing zeroes in n!	⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...
Unit digit cyclicity (length 4)	2→2,4,8,6; 3→3,9,7,1; 7→7,9,3,1; 8→8,4,2,6
Unit digit cyclicity (length 2)	4→4,6; 9→9,1
Unit digit (length 1)	0→0; 1→1; 5→5; 6→6
Euler's theorem	a^φ(n) ≡ 1 (mod n) for gcd(a,n)=1
Fermat's Little Theorem	a^(p−1) ≡ 1 (mod p) for prime p
Wilson's theorem (prime p)	(p−1)! ≡ −1 (mod p)
Digital root (divisibility by 9)	Sum of digits reduced to single digit
Square of number ending in 5	(a5)² = a(a+1) × 100 + 25
Square root approximation	√N ≈ (N + P²) / (2P) where P² is nearest perfect square

Previous Years' Trend Analysis (2020–2025)

Analysis of the last 5 years of SSC CGL papers reveals the following patterns:

Year	Tier	Question Type	Difficulty
2020	Tier 1	Divisibility by 8&9 combined, Unit digit of sum of powers	Easy
2020	Tier 2	Number of factors, Remainder using Euler theorem	Medium
2021	Tier 1	LCM-HCF product formula, Divisibility by 11	Easy
2021	Tier 2	HCF of fractions, Surd comparison, Base conversion	Hard
2022	Tier 1	Unit digit of factorial sum, Composite divisibility	Easy
2022	Tier 2	Wilson's theorem application, Number of even/odd factors	Hard
2023	Tier 1	BODMAS simplification, Finding prime numbers	Easy
2023	Tier 2	Remainder with negative concept, Cyclicity of large powers	Medium
2024	Tier 1	LCM of three numbers, Divisibility rule for 7	Easy
2024	Tier 2	Finding missing digit for divisibility by 9&11, Advanced surd rationalization	Medium
2025	Tier 1	Unit digit of 3⁴²+7²³, HCF using difference method	Easy

Key Takeaways: Divisibility rules appear in nearly every exam. Unit digit questions are the most frequent individual question type. LCM-HCF questions have shifted towards fraction forms in recent years. Remainder theorem questions appear more in Tier 2 than Tier 1.

Quick Reference: Divisibility at a Glance

Divisor	Rule	Example
2	Last digit even (0,2,4,6,8)	1234 → 4 → divisible
3	Sum of digits divisible by 3	123 → 1+2+3=6 → divisible
4	Last two digits divisible by 4	7324 → 24 → divisible
5	Last digit 0 or 5	1235 → 5 → divisible
6	Divisible by both 2 and 3	1236 → even, sum=12 → divisible
7	Double last digit, subtract from rest	343 → 34−6=28 → divisible
8	Last three digits divisible by 8	73128 → 128 → divisible
9	Sum of digits divisible by 9	12345 → 15 → not divisible
10	Last digit 0	12340 → divisible
11	Alternate sum difference = 0 or multiple of 11	13321 → 5−5=0 → divisible

Additional Practice: 10 Quick Warm-up Questions

Solve these before attempting the main practice set below. Each should take under 15 seconds.

Is 567 divisible by 9? (Hint: sum of digits)
What is the unit digit of 8²³?
Find the HCF of 42 and 56 using the difference method.
How many factors does 100 have?
What is the remainder when 5¹⁰ is divided by 5?
Is 7,11,13 divisibility: Is 1001 divisible by 7?
What is the LCM of 8, 12, and 18?
Find the square of 75 using the Vedic trick.
Which is larger: √8 or ∛20?
Simplify: 48 ÷ 8 × 2 + 6 − 4

Answers: 1: Yes (sum=18), 2: 2 (cyclicity 8,4,2,6; 23 rem 3 = 2), 3: 14, 4: 9 (100=2²×5²), 5: 0, 6: Yes (1001=7×11×13), 7: 72, 8: 5625, 9: √8 (LCM 6: 512 vs 400), 10: 8

Practice Questions

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