Number Relationships and Square Roots

Number Relationships and Square Roots is a high-scoring chapter for SSC CGL. Every year 1 to 2 questions come directly from this chapter. This chapter covers the relationship between numbers, properties of numbers, HCF and LCM concepts, remainders, and complete coverage of square roots and cube roots. Students who understand the relationships between numbers can solve problems much faster using shortcuts rather than lengthy calculations.

1. Relationships Between Numbers

1.1 Factors and Multiples - Deep Coverage

Factor:

• a is a factor of b if b is completely divisible by a (remainder = 0)
• In other words, b = a × k for some integer k
• Example: Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Properties of Factors:

• 1 is a factor of every number
• Every number is a factor of itself
• A factor of a number is always less than or equal to the number
• If a is a factor of b and b is a factor of c, then a is a factor of c (transitivity)
• If a is a factor of b and a is a factor of c, then a is a factor of (b + c) and (b - c)
• The smallest factor of any number greater than 1 is always a prime number

Multiple:

• b is a multiple of a if a is a factor of b
• Multiples of 5: 5, 10, 15, 20, 25, 30, ...
• Every number is a multiple of 1
• Every number is a multiple of itself
• Multiples of a number are infinite

Finding all factors of a number: 

Step 1: Find prime factorisation 

Step 2: Use formula (a+1)(b+1)(c+1)... to get count 

Step 3: List all factors systematically in pairs

Example: Find all factors of 60 

60 = 2² × 3 × 5 

Number of factors = (2+1)(1+1)(1+1) = 3 × 2 × 2 = 12

Factors in pairs: (1, 60), (2, 30), (3, 20), (4, 15), (5, 12), (6, 10) 

All factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Sum of all factors formula: If N = p^a × q^b × r^c then: Sum of factors = [(p^(a+1) - 1)/(p-1)] × [(q^(b+1) - 1)/(q-1)] × [(r^(c+1) - 1)/(r-1)]

Example: Sum of factors of 12 

12 = 2² × 3 

Sum = [(2³-1)/(2-1)] × [(3²-1)/(3-1)] 

= [7/1] × [8/2] 

= 7 × 4 = 28 

Verify: 1+2+3+4+6+12 = 28 ✓

Product of all factors: If N has n factors, product of all factors = N^(n/2)

Example: Product of all factors of 12 12 has 6 factors. Product = 12^(6/2) = 12³ = 1728 Verify: 1×2×3×4×6×12 = 1728 ✓

1.2 HCF (Highest Common Factor)

Definition: The HCF of two or more numbers is the largest number that divides all of them exactly. Also called GCD (Greatest Common Divisor).

Method 1: Prime Factorisation Step 1: Express each number as a product of prime factors Step 2: Take the common prime factors Step 3: Choose the LOWEST power of each common prime factor Step 4: Multiply them

Example: HCF of 36 and 48 

36 = 2² × 3² 

48 = 2⁴ × 3 Common primes: 2 and 3 

HCF = 2² × 3¹ = 4 × 3 = 12

Example: HCF of 72, 108 and 144 

72 = 2³ × 3² 

108 = 2² × 3³ 

144 = 2⁴ × 3² 

HCF = 2² × 3² = 4 × 9 = 36

Method 2: Division Method (Euclidean Algorithm) Step 1: Divide larger number by smaller Step 2: Divide previous divisor by remainder Step 3: Repeat until remainder = 0 Step 4: Last non-zero divisor is the HCF

Example: HCF of 48 and 36 

48 ÷ 36 = 1 remainder 12 

36 ÷ 12 = 3 remainder 0 

HCF = 12

Example: HCF of 252 and 105 

252 ÷ 105 = 2 remainder 42 

105 ÷ 42 = 2 remainder 21 

42 ÷ 21 = 2 remainder 0 

HCF = 21

Properties of HCF:

• HCF(a, b) ≤ min(a, b)
• HCF(a, b) is a factor of both a and b
• HCF(a, b) = HCF(b, a) (commutative)
• HCF(a, a) = a
• If a divides b, then HCF(a, b) = a
• HCF(a, b) × LCM(a, b) = a × b (for two numbers)

1.3 LCM (Least Common Multiple)

Definition: The LCM of two or more numbers is the smallest number that is exactly divisible by all of them.

Method 1: Prime Factorisation Step 1: Express each number as product of prime factors Step 2: Take ALL prime factors that appear in any number Step 3: Choose the HIGHEST power of each prime factor Step 4: Multiply them

Example: LCM of 36 and 48 

36 = 2² × 3² 

48 = 2⁴ × 3 

LCM = 2⁴ × 3² = 16 × 9 = 144

Example: LCM of 12, 18 and 24 

12 = 2² × 3 

18 = 2 × 3² 

24 = 2³ × 3 

LCM = 2³ × 3² = 8 × 9 = 72

Method 2: Division Method Divide all numbers simultaneously by common prime factors until all become 1. LCM = product of all divisors used.

Example: LCM of 12, 18, 24 

2 | 12, 18, 24 

2 | 6, 9, 12 

2 | 3, 9, 6 

3 | 3, 9, 3 

3 | 1, 3, 1 

| 1, 1, 1 

LCM = 2 × 2 × 2 × 3 × 3 = 72

Properties of LCM:

• LCM(a, b) ≥ max(a, b)
• LCM(a, b) is a multiple of both a and b
• LCM(a, b) = LCM(b, a)
• LCM(a, a) = a
• If a divides b, then LCM(a, b) = b
• HCF × LCM = product of the two numbers (for two numbers only)

1.4 Relationship Between HCF and LCM

Key Formula: HCF(a, b) × LCM(a, b) = a × b

This works ONLY for two numbers. For three or more numbers: HCF × LCM ≠ product of numbers (in general)

Important results:

• If HCF of two numbers is H, both numbers can be written as Ha and Hb where HCF(a,b) = 1
• LCM = H × a × b
• So LCM/HCF = a × b (product of co-prime quotients)

Example: HCF of two numbers is 6 and LCM is 72. One number is 18. Find the other. Using HCF × LCM = product of numbers: 6 × 72 = 18 × other Other = 6 × 72/18 = 432/18 = 24

Relationship for fractions:

• HCF of fractions = HCF of numerators / LCM of denominators
• LCM of fractions = LCM of numerators / HCF of denominators

Example: HCF of 3/4 and 5/6 

= HCF(3,5) / LCM(4,6) 

= 1/12

Example: LCM of 3/4 and 5/6 

= LCM(3,5) / HCF(4,6) 

= 15/2

1.5 Important SSC Problems on HCF and LCM

Type 1: Finding numbers given HCF and LCM

If HCF = H and LCM = L, then numbers are of form Ha and Hb where ab = L/H and HCF(a,b) = 1.

Example: HCF = 12, LCM = 144. Find possible pairs of numbers. 

L/H = 144/12 = 12 = 1 × 12 or 3 × 4 (co-prime pairs) 

Pairs: (12×1, 12×12) = (12, 144) or (12×3, 12×4) = (36, 48)

Type 2: Largest number that divides given numbers leaving same remainder

"Find the largest number that divides a, b, c leaving the same remainder" = HCF of (a-b), (b-c), (a-c)

Example: Find largest number that divides 77, 123, 169 leaving the same remainder. 

123 - 77 = 46 

169 - 123 = 46 

169 - 77 = 92 

HCF(46, 46, 92) = 46

Type 3: Largest number that divides given numbers leaving specific remainders

"Find largest number that divides a leaving remainder r₁ and divides b leaving remainder r₂" = HCF(a - r₁, b - r₂)

Example: Find largest number that divides 77 leaving remainder 5 and divides 123 leaving remainder 3. 

= HCF(77-5, 123-3) = HCF(72, 120) 

72 = 2³ × 3² 

120 = 2³ × 3 × 5 

HCF = 2³ × 3 = 24

Type 4: Smallest number divisible by given numbers = LCM of given numbers

Type 5: Smallest number that when divided by given numbers leaves remainder r = LCM of given numbers + r

Example: Smallest number that when divided by 4, 6, and 9 leaves remainder 3 in each case. 

LCM(4, 6, 9) = 36 

Answer = 36 + 3 = 39 

Verify: 39 ÷ 4 = 9 remainder 3 ✓, 39 ÷ 6 = 6 remainder 3 ✓, 39 ÷ 9 = 4 remainder 3 ✓

Type 6: Largest number that divides numbers exactly (no remainder) = HCF of the numbers

Type 7: Number of common factors of two numbers = Number of factors of HCF of those two numbers

2. Remainder Theorem and Related Concepts

2.1 Division Algorithm

For any two positive integers a (dividend) and b (divisor): a = b × q + r where 0 ≤ r < b

Here q = quotient and r = remainder

Example: 47 = 5 × 9 + 2 (dividend = 47, divisor = 5, quotient = 9, remainder = 2)

Key relationships:

• Dividend = Divisor × Quotient + Remainder
• Quotient = (Dividend - Remainder) / Divisor
• Remainder = Dividend - Divisor × Quotient

2.2 Properties of Remainders

Property 1: When a number N is divided by d, remainder = r Then kN divided by d gives remainder = remainder of (k × r) ÷ d

Property 2: If N₁ divided by d gives r₁ and N₂ divided by d gives r₂, then:

• (N₁ + N₂) divided by d gives remainder = remainder of (r₁ + r₂) ÷ d
• (N₁ × N₂) divided by d gives remainder = remainder of (r₁ × r₂) ÷ d

Property 3: Remainder when N is divided by d = N - d × floor(N/d)

2.3 Remainder Shortcuts for SSC

Remainder when divided by 9: Remainder = sum of digits (reduce until single digit if needed) Example: 347 ÷ 9: sum = 3+4+7 = 14, 1+4 = 5. Remainder = 5. Verify: 347 = 9 × 38 + 5 ✓

Remainder when divided by 3: Remainder = remainder when sum of digits is divided by 3 Example: 347 ÷ 3: sum = 14, 14 ÷ 3 = 4 remainder 2. Remainder = 2.

Remainder when divided by 10: Remainder = last digit of the number Example: 347 ÷ 10 = Remainder 7

Remainder when divided by 5: Remainder = last digit mod 5 Example: 347 ÷ 5: last digit 7, 7 mod 5 = 2. Remainder = 2.

Remainder when divided by 2: Remainder = last digit mod 2 (0 if even, 1 if odd)

Remainder when 10^n is divided by 9: Always = 1 (since 10 ≡ 1 mod 9, so 10^n ≡ 1 mod 9)

Remainder when 2^n is divided by different numbers:

2.4 Cyclicity of Last Digits (Power Cycles)

Last digit of powers follows a cycle:

How to find last digit of a^n: 

Step 1: Find the cycle of last digit of base a 

Step 2: Divide n by the period of the cycle 

Step 3: The remainder tells which element of cycle is the answer 

Step 4: If remainder = 0, take last element of cycle

Example: Last digit of 7^45 

Cycle of 7: 7, 9, 3, 1 (period 4) 

45 ÷ 4 = 11 remainder 1 

Position 1 in cycle = 7 

Last digit of 7^45 = 7

Example: Last digit of 3^100 

Cycle of 3: 3, 9, 7, 1 (period 4) 

100 ÷ 4 = 25 remainder 0 

Remainder 0 means last element of cycle = 1 

Last digit of 3^100 = 1

Example: Last digit of 2^50 

Cycle of 2: 2, 4, 8, 6 (period 4) 

50 ÷ 4 = 12 remainder 2 

Position 2 in cycle = 4 

Last digit of 2^50 = 4

3. Square Roots - Complete Deep Coverage

3.1 Definition and Basic Properties

Definition: If b² = a, then b is called the square root of a, written as √a = b (b ≥ 0)

The positive square root is called the principal square root.

Fundamental properties:

• √(a × b) = √a × √b (for a, b ≥ 0)
• √(a/b) = √a/√b (for a ≥ 0, b > 0)
• (√a)² = a (for a ≥ 0)
• √(a²) = |a| (equals a if a ≥ 0, equals -a if a < 0)
• √a × √a = a
• √a + √b ≠ √(a+b) — this is a very common error

Condition for √a to be real: a must be ≥ 0. √(-1) = i (imaginary, not real)

3.2 Perfect Squares - Complete List and Properties

Perfect squares from 1 to 400: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400

Their square roots: √1=1, √4=2, √9=3, √16=4, √25=5, √36=6, √49=7, √64=8, √81=9, √100=10 √121=11, √144=12, √169=13, √196=14, √225=15, √256=16, √289=17, √324=18, √361=19, √400=20

Properties of perfect squares:

• A perfect square always ends in: 0, 1, 4, 5, 6, 9
• A perfect square NEVER ends in: 2, 3, 7, 8
• A perfect square ending in 1 has square root ending in 1 or 9
• A perfect square ending in 4 has square root ending in 2 or 8
• A perfect square ending in 5 has square root ending in 5
• A perfect square ending in 6 has square root ending in 4 or 6
• A perfect square ending in 9 has square root ending in 3 or 7
• A perfect square ending in 0 has square root ending in 0 (and actually ends in 00)
• The square of an odd number is always odd
• The square of an even number is always even
• If n is odd: n² ≡ 1 (mod 8) [odd perfect squares leave remainder 1 when divided by 8]
• Sum of first n odd numbers = n²

Difference of consecutive squares: n² - (n-1)² = 2n - 1 (always an odd number) 

Examples: 4² - 3² = 7, 5² - 4² = 9, 10² - 9² = 19

3.3 Methods to Find Square Root

Method 1: Prime Factorisation 

Step 1: Express number as product of prime factors 

Step 2: Group factors in pairs 

Step 3: Take one from each pair and multiply

Example: √1764 

1764 = 2² × 3² × 7² 

= (2 × 3 × 7)² 

= 42² √1764 

= 42

Example: √11025 

11025 = 3² × 5² × 7² 

= (3 × 5 × 7)² 

= 105² √11025 

= 105

Method 2: Long Division Method Used for large numbers or non-perfect squares.

Steps:

1. Group digits in pairs from right (decimal: group from decimal point both ways)
2. Find largest integer whose square ≤ first group
3. Subtract and bring down next pair
4. Double current result for trial divisor
5. Find digit d such that (20×current + d) × d ≤ current remainder
6. Repeat

Example: √784 

Group: 7 | 84 

√7 ≈ 2 (since 2² = 4 ≤ 7 < 9 = 3²) 

2² = 4, remainder = 7-4 = 3 

Bring down 84: 384 

Trial divisor = 2 × 2 = 4_ Find d: 48 × 8 = 384 ✓ √784 = 28

Method 3: Approximation using nearby perfect squares For non-perfect squares, find √n by locating between consecutive perfect squares.

Example: Approximate √50 

√49 = 7 and √64 = 8 

50 is closer to 49 than to 64 

√50 ≈ 7.07 (actual: 7.071...)

Better approximation: √50 = √(49+1) ≈ 7 + 1/(2×7) = 7 + 1/14 ≈ 7.07

Method 4: Digit-by-digit estimation

The number of digits in √n:

• If n has odd number of digits: √n has (digits of n + 1)/2 digits
• If n has even number of digits: √n has (digits of n)/2 digits

Example: √5476 → 4 digits → √ has 2 digits → answer between 70 and 80 

Check 74² = 5476 ✓

3.4 Simplifying Square Roots

Finding simplified form of √n: 

Step 1: Find the largest perfect square factor of n 

Step 2: √n = √(perfect square × remaining) = √(perfect square) × √(remaining)

Examples:

• √72 = √(36 × 2) = 6√2
• √50 = √(25 × 2) = 5√2
• √108 = √(36 × 3) = 6√3
• √180 = √(36 × 5) = 6√5
• √288 = √(144 × 2) = 12√2
• √245 = √(49 × 5) = 7√5
• √363 = √(121 × 3) = 11√3
• √500 = √(100 × 5) = 10√5
• √1800 = √(900 × 2) = 30√2

Simplifying expressions: 

√(a²b) = a√b (for a > 0) 

√(a²b²c) = ab√c

3.5 Square Roots of Fractions and Decimals

Square root of a fraction: 

√(p/q) = √p/√q

Examples:

• √(9/16) = 3/4
• √(25/49) = 5/7
• √(4/9) = 2/3
• √(121/169) = 11/13

Square root of a decimal: 

Convert to fraction or handle carefully.

Examples:

• √0.25 = √(25/100) = 5/10 = 0.5
• √0.09 = √(9/100) = 3/10 = 0.3
• √0.0049 = √(49/10000) = 7/100 = 0.07
• √1.44 = √(144/100) = 12/10 = 1.2
• √2.25 = √(225/100) = 15/10 = 1.5

Pattern for decimals:

• √0.__ (two decimal places) = 0._ (one decimal place) if perfect square
• √0.____ (four decimal places) = 0.__ (two decimal places) if perfect square

3.6 Square Roots of Large Numbers - Tricks

Trick 1: Finding square root ending digit

Given the last two digits of a perfect square, determine the possible last digit of its square root:

Trick 2: Quick check method For a 4-digit perfect square N:

• √N is a 2-digit number
• Find a: a² ≤ first two digits of N
• Find b: the last digit based on last digit of N
• Check if (10a + b)² = N

Example: √5329 

First two digits: 53. a² ≤ 53 → a = 7 (since 7² = 49 ≤ 53 < 64 = 8²) 

Last digit of 5329 = 9 → last digit of root = 3 or 7 

Check 73² = 5329 ✓

3.7 Irrational Square Roots - Values to Memorise

Essential values for SSC:

√2 = 1.4142135... 

√3 = 1.7320508... 

√5 = 2.2360679... 

√6 = 2.4494897... 

√7 = 2.6457513... 

√8 = 2√2 = 2.8284271... 

√10 = 3.1622776... 

√11 = 3.3166247... 

√12 = 2√3 = 3.4641016... 

√13 = 3.6055512... 

√14 = 3.7416573... 

√15 = 3.8729833... 

√17 = 4.1231056... 

√18 = 3√2 = 4.2426406... 

√19 = 4.3588989... √20 = 2√5 = 4.4721359...

Useful approximations:

• 1/√2 = √2/2 ≈ 0.707
• 1/√3 = √3/3 ≈ 0.577
• 2/√3 = 2√3/3 ≈ 1.155

3.8 Operations on Square Roots

Addition and Subtraction (like surds only): a√n ± b√n = (a ± b)√n

Examples:

• 3√2 + 5√2 = 8√2
• 7√3 - 4√3 = 3√3
• 2√5 + √5 = 3√5

Unlike surds cannot be directly added: 

√2 + √3 ≠ √5 (common mistake)

But we can simplify by making them like surds: 

√8 + √18 = 2√2 + 3√2 = 5√2

More examples:

• √12 + √27 = 2√3 + 3√3 = 5√3
• √50 - √8 = 5√2 - 2√2 = 3√2
• √45 + √20 = 3√5 + 2√5 = 5√5
• √72 - √50 = 6√2 - 5√2 = √2

Multiplication: √a × √b = √(ab) Examples:

• √3 × √12 = √36 = 6
• √5 × √20 = √100 = 10
• 2√3 × 3√5 = 6√15
• √(2+√3) × √(2-√3) = √(4-3) = √1 = 1

Division: √a/√b = √(a/b) Examples:

• √12/√3 = √4 = 2
• √75/√3 = √25 = 5
• 6√15/3√5 = 2√3

3.9 Rationalisation - Complete Coverage

Type 1: Single surd in denominator Multiply numerator and denominator by the same surd.

• 5/√3 = 5√3/3
• 3/√7 = 3√7/7
• a/√b = a√b/b

Type 2: Binomial denominator with one surd (a + √b) Multiply by conjugate (a - √b).

• 1/(5 + √3) × (5-√3)/(5-√3) = (5-√3)/(25-3) = (5-√3)/22
• 1/(√5 + √3) × (√5-√3)/(√5-√3) = (√5-√3)/2

Type 3: Binomial denominator with two surds (√a + √b) Multiply by conjugate (√a - √b).

• 1/(√7 - √5) × (√7+√5)/(√7+√5) = (√7+√5)/(7-5) = (√7+√5)/2
• 1/(√11 + √7) = (√11-√7)/(11-7) = (√11-√7)/4

Type 4: More complex denominators May need to rationalise twice.

(1+√2)/(1-√2) = (1+√2)²/((1-√2)(1+√2)) = (1+2√2+2)/(1-2) = (3+2√2)/(-1) = -(3+2√2)

3.10 Important Surd Expressions for SSC

Pattern 1: (√a + √b)(√a - √b) = a - b

• (√5 + √3)(√5 - √3) = 5-3 = 2
• (√11 + √7)(√11 - √7) = 11-7 = 4
• (3 + √2)(3 - √2) = 9-2 = 7

Pattern 2: (√a + √b)² = a + 2√(ab) + b

• (√3 + √2)² = 3 + 2√6 + 2 = 5 + 2√6
• (√5 + √3)² = 5 + 2√15 + 3 = 8 + 2√15

Pattern 3: Nested square roots √(a + 2√b) = √c + √d where c + d = a and cd = b

Example: √(7 + 2√12) = √(7 + 4√3) 

Looking for √c + √d where c+d=7 and cd=12 

c=4, d=3: √4 + √3 = 2 + √3 

Verify: (2+√3)² = 4 + 4√3 + 3 = 7 + 4√3 ✓ 

So √(7 + 4√3) = 2 + √3

Example: √(6 - 2√5) 

= √(5 - 2√5 + 1) = √(√5 - 1)² = √5 - 1

Pattern 4: Continued square roots 

x = √(n + √(n + √(n + ...))) 

x = √(n + x) → x² = n + x → x² - x - n = 0 

x = (1 + √(1+4n))/2

Example: x = √(2 + √(2 + √(2 + ...))) x² - x - 2 = 0 → (x-2)(x+1) = 0 → x = 2

4. Cube Roots - Complete Coverage

4.1 Perfect Cubes - Complete List

Perfect cubes from 1³ to 20³:

4.2 Properties of Perfect Cubes

Last digit pattern of cubes:

Key observation: The last digit of a cube uniquely determines the last digit of its cube root (unlike squares). This is extremely useful for SSC.

From the table:

• Cube ending in 1 → cube root ends in 1
• Cube ending in 8 → cube root ends in 2
• Cube ending in 7 → cube root ends in 3
• Cube ending in 4 → cube root ends in 4
• Cube ending in 5 → cube root ends in 5
• Cube ending in 6 → cube root ends in 6
• Cube ending in 3 → cube root ends in 7
• Cube ending in 2 → cube root ends in 8
• Cube ending in 9 → cube root ends in 9
• Cube ending in 0 → cube root ends in 0

4.3 Finding Cube Root of Large Perfect Cubes (SSC Trick)

Method for 5-6 digit perfect cubes:

Step 1: Group digits in sets of 3 from right 

Step 2: Find cube root of right group → gives last digit of answer 

Step 3: Find cube root of left group (find n such that n³ ≤ left group < (n+1)³) → gives first digit

Example: ∛17576 

Right group: 576 → ends in 6 → cube root ends in 6 

Left group: 17 → 2³=8 ≤ 17 < 27=3³ → first digit is 2 

Answer = 26 

Verify: 26³ = 17576 ✓

Example: ∛32768 

Right group: 768 → ends in 8 → cube root ends in 2 

Left group: 32 → 3³=27 ≤ 32 < 64=4³ → first digit is 3 

Answer = 32 

Verify: 32³ = 32768 ✓

Example: ∛314432 

Right group: 432 → ends in 2 → cube root ends in 8 

Left group: 314 → 6³=216 ≤ 314 < 343=7³ → first digit is 6 

Answer = 68 

Verify: 68³ = 314432 ✓

4.4 Properties of Cube Roots

• ∛(a × b) = ∛a × ∛b
• ∛(a/b) = ∛a/∛b
• (∛a)³ = a
• ∛(a³) = a
• ∛(-a) = -∛a (cube roots of negative numbers are real and negative)
• ∛(-8) = -2, ∛(-27) = -3, ∛(-125) = -5

Difference from square roots: Cube root of a negative number IS real (unlike square root of negative).

5. Number Relationships - Special Types

5.1 Special Number Pairs

Amicable Numbers: Pairs where each is the sum of proper divisors of the other.

• (220, 284): Proper divisors of 220 sum to 284 and vice versa

Consecutive numbers:

• Product of consecutive numbers is always divisible by 2
• Product of any 3 consecutive numbers is divisible by 6
• Product of any 4 consecutive numbers is divisible by 24
• Product of any n consecutive numbers is divisible by n!

Key results for consecutive integers:

n(n+1) is always even (one of n, n+1 is even) n(n+1)(n+2) is always divisible by 6 n(n+1)(2n+1) is always divisible by 6

5.2 Divisibility by Special Numbers

Divisibility of a^n - b^n:

• a^n - b^n is always divisible by (a - b)
• If n is even: a^n - b^n is divisible by (a + b) also
• If n is odd: a^n + b^n is divisible by (a + b)

Key SSC application:

• 10^n - 1 = 999...9 (n nines) is divisible by 9
• 10^n - 1 is divisible by 9, 99, 999... and their factors
• 2^10 = 1024, so 2^10 - 1 = 1023 = 3 × 341

5.3 Number Patterns for SSC

Sum of n terms of AP (Arithmetic Progression): S = n/2 × (first term + last term) = n/2 × (2a + (n-1)d) where a = first term, d = common difference

Useful results:

• 1 + 2 + 3 + ... + n = n(n+1)/2
• 2 + 4 + 6 + ... + 2n = n(n+1)
• 1 + 3 + 5 + ... + (2n-1) = n²
• a² - b² = (a+b)(a-b)
• a³ - b³ = (a-b)(a² + ab + b²)
• a³ + b³ = (a+b)(a² - ab + b²)

5.4 Indices and Exponents

Laws of Indices:

• a^m × a^n = a^(m+n)
• a^m ÷ a^n = a^(m-n)
• (a^m)^n = a^(mn)
• (ab)^n = a^n × b^n
• (a/b)^n = a^n/b^n
• a^0 = 1 (for a ≠ 0)
• a^(-n) = 1/a^n
• a^(1/n) = ⁿ√a
• a^(m/n) = (ⁿ√a)^m = ⁿ√(a^m)

Special cases:

• 0^0 = undefined
• 0^n = 0 for n > 0
• 1^n = 1 for all n
• (-1)^n = 1 if n is even, -1 if n is odd

Comparison of powers: To compare a^m and b^n:

• Express both with same base, or
• Express both with same exponent, or
• Convert to decimals

Example: Compare 2^30 and 3^20 

2^30 = (2^3)^10 = 8^10 

3^20 = (3^2)^10 = 9^10 

Since 9 > 8, we have 3^20 > 2^30

6. Solved Examples - SSC Level

Example 1: Find HCF and LCM of 36 and 48. 

36 = 2² × 3², 48 = 2⁴ × 3 

HCF = 2² × 3 = 12 

LCM = 2⁴ × 3² = 144 

Verify: HCF × LCM = 12 × 144 = 1728 = 36 × 48 ✓

Example 2: The HCF of two numbers is 8 and their LCM is 192. If one number is 64, find the other. 

HCF × LCM = product of numbers 

8 × 192 = 64 × other 

Other = 1536/64 = 24

Example 3: Find the largest number that divides 625, 761 and 999 leaving the same remainder. 

761 - 625 = 136 

999 - 761 = 238 

999 - 625 = 374 

HCF(136, 238, 374): 

136 = 2³ × 17 

238 = 2 × 7 × 17 

374 = 2 × 11 × 17 

HCF = 2 × 17 = 34

Example 4: Find the smallest number that when divided by 8, 12 and 15 leaves remainder 5. 

LCM(8, 12, 15) = 120 

Answer = 120 + 5 = 125 

Verify: 125 ÷ 8 = 15 remainder 5 ✓, 125 ÷ 12 = 10 remainder 5 ✓, 125 ÷ 15 = 8 remainder 5 ✓

Example 5: Find the last digit of 7^55. 

Cycle of 7: 7, 9, 3, 1 (period 4) 

55 ÷ 4 = 13 remainder 3 

Position 3 = 3 

Last digit of 7^55 = 3

Example 6: Find ∛474552. 

Right group: 552 → ends in 2 → cube root ends in 8 

Left group: 474 → 7³=343 ≤ 474 < 512=8³ → first digit is 7 

Answer = 78 

Verify: 78³ = 78 × 78 × 78 = 6084 × 78 = 474552 ✓

Example 7: Simplify √(8 + 2√15). 

Looking for (√a + √b)² = a + b + 2√(ab) = 8 + 2√15 

So a + b = 8 and ab = 15 

a = 5, b = 3 (since 5+3=8 and 5×3=15) 

√(8 + 2√15) = √5 + √3

Example 8: If HCF of two numbers is 12 and their sum is 84, find the LCM. 

Numbers = 12a and 12b where HCF(a,b) = 1 

12a + 12b = 84 → a + b = 7 

Since HCF(a,b) = 1 and a+b = 7: 

Possible pairs (a,b): (1,6), (2,5), (3,4) 

But HCF(2,5) = 1 ✓, HCF(3,4) = 1 ✓, HCF(1,6) = 1 ✓ 

LCM = 12 × a × b: 

For (1,6): LCM = 12 × 6 = 72, numbers = (12, 72) 

For (2,5): LCM = 12 × 10 = 120, numbers = (24, 60) 

For (3,4): LCM = 12 × 12 = 144, numbers = (36, 48)

Example 9: Find √(√(√65536)). 

65536 = 2^16 

√65536 = 2^8 = 256 

√256 = 2^4 = 16 

√16 = 2^2 = 4 

Answer = 4

Example 10: Find the value of √(12 + √(12 + √(12 + ...∞))) 

Let x = √(12 + x) 

x² = 12 + x 

x² - x - 12 = 0 

(x - 4)(x + 3) = 0 

x = 4 (taking positive value)

7. Key Formulas and Quick Reference

HCF and LCM:

• HCF × LCM = product of two numbers
• HCF of fractions = HCF of numerators / LCM of denominators
• LCM of fractions = LCM of numerators / HCF of denominators
• Largest number dividing a, b, c with same remainder = HCF(a-b, b-c, a-c)
• Smallest number divisible by a, b, c with remainder r = LCM(a,b,c) + r

Square Roots:

• Sum of first n odd numbers = n²
• (n+1)² - n² = 2n+1 (difference of consecutive squares)
• √a + √b: cannot simplify unless like surds
• √(a+2√b) = √c + √d where c+d = a, cd = b
• x = √(n + x) gives x = (1 + √(1+4n))/2

Cube Roots (Last digit trick):

• Ends 1 → root ends 1
• Ends 8 → root ends 2
• Ends 7 → root ends 3
• Ends 4 → root ends 4
• Ends 5 → root ends 5
• Ends 6 → root ends 6
• Ends 3 → root ends 7
• Ends 2 → root ends 8
• Ends 9 → root ends 9
• Ends 0 → root ends 0

Cyclicity of last digits:

• Last digit 2, 3, 7, 8: cycle of period 4
• Last digit 4, 9: cycle of period 2
• Last digit 1, 5, 6, 0: always same (period 1)

Indices:

• a^m × a^n = a^(m+n)
• (a^m)^n = a^(mn)
• a^(1/n) = ⁿ√a
• a^(-n) = 1/a^n