Number System Tricks: Master HCF, LCM, Divisibility and Remainders Fast

Number system is the one topic in competitive mathematics where a single well-memorized trick can save 60–90 seconds per question. Yet it is also the topic most students prepare least systematically — they learn divisibility rules in isolation, pick up a remainder shortcut here and there, and never develop a coherent mental framework for the full topic.

The result: avoidable errors on questions that should be among the fastest to solve.

This guide builds that coherent framework. It covers every major number system sub-topic tested in SSC CGL, RRB NTPC, IBPS PO, TCS NQT, and school board examinations — divisibility, HCF, LCM, unit digits, remainders, factor counting, and consecutive number properties — with the fastest available method for each type and multiple worked examples throughout.

Part 1: Divisibility Rules — Complete Reference

Divisibility rules are the entry point for almost every number system problem. Knowing them instantly eliminates calculation and enables rapid factor identification.

The Complete Divisibility Rule Set

Divisible by 2: Last digit is 0, 2, 4, 6, or 8

• 4,738 ✓ | 5,931 ✗

Divisible by 3: Sum of digits divisible by 3

• 4,731: 4+7+3+1 = 15 → 15÷3 = 5 ✓
• 4,735: 4+7+3+5 = 19 → 19÷3 has remainder 1 ✗

Divisible by 4: Last two digits divisible by 4

• 4,732: 32÷4 = 8 ✓
• 4,734: 34÷4 = 8.5 ✗

Divisible by 5: Last digit is 0 or 5

• 4,735 ✓ | 4,733 ✗

Divisible by 6: Divisible by BOTH 2 and 3

• 4,734: even ✓ and digit sum = 18 ✓ → divisible by 6 ✓

Divisible by 7: Double the last digit, subtract from remaining number. Repeat until manageable.

• 343: 34 − (3×2) = 34 − 6 = 28 → 28÷7 = 4 ✓
• 532: 53 − (2×2) = 53 − 4 = 49 → 49÷7 = 7 ✓
• 678: 67 − (8×2) = 67 − 16 = 51 → 51÷7 = 7.28 ✗

Divisible by 8: Last three digits divisible by 8

• 4,736: 736÷8 = 92 ✓
• 4,734: 734÷8 = 91.75 ✗

Divisible by 9: Sum of digits divisible by 9

• 4,734: 4+7+3+4 = 18 → 18÷9 = 2 ✓
• 4,731: digit sum = 15 → 15÷9 has remainder 6 ✗

Divisible by 10: Last digit is 0

• 4,730 ✓

Divisible by 11: Difference between sum of digits at odd positions and sum at even positions is 0 or divisible by 11

• 4,741: Odd positions (1st, 3rd) = 4+4 = 8, Even positions (2nd, 4th) = 7+1 = 8 → difference = 0 ✓
• 5,863: Odd = 5+6 = 11, Even = 8+3 = 11 → difference = 0 ✓
• 9,427: Odd = 9+2 = 11, Even = 4+7 = 11 → difference = 0 ✓

Divisible by 12: Divisible by BOTH 3 and 4

• 4,728: digit sum = 21 ✓ (÷3), last two digits 28÷4 = 7 ✓ → divisible by 12 ✓

Divisible by 25: Last two digits are 00, 25, 50, or 75

• 4,725 ✓ | 4,735 ✗

Part 2: HCF — Three Methods, One Best Choice

Method 1: Prime Factorization

Factorize both numbers, take the product of common prime factors with lowest powers.

Example: HCF of 72 and 120

• 72 = 2³ × 3²
• 120 = 2³ × 3 × 5
• HCF = 2³ × 3 = 24

Best for: Numbers under 100 with obvious prime factors.

Method 2: Euclidean Algorithm (Fastest for Large Numbers)

Divide the larger by the smaller, replace the larger with the remainder, repeat until remainder = 0.

Example: HCF of 252 and 336

• 336 = 1 × 252 + 84
• 252 = 3 × 84 + 0
• HCF = 84

Example: HCF of 315 and 525

• 525 = 1 × 315 + 210
• 315 = 1 × 210 + 105
• 210 = 2 × 105 + 0
• HCF = 105

Best for: Any two numbers — the most universally reliable method.

Method 3: Subtraction Method (For Close Numbers)

Repeatedly subtract the smaller from the larger until both are equal.

Example: HCF of 88 and 99

• 99 − 88 = 11
• HCF(88, 11): 88 = 8 × 11 → HCF = 11

Best for: Numbers that differ by a small amount.

Part 3: LCM — The Fastest Methods

Method 1: LCM = Product ÷ HCF

LCM(a, b) = (a × b) ÷ HCF(a, b)

Example: LCM of 36 and 48

• HCF(36, 48) = 12
• LCM = 36 × 48 ÷ 12 = 36 × 4 = 144

Example: LCM of 84 and 120

• HCF(84, 120): 120 = 1×84+36, 84 = 2×36+12, 36 = 3×12+0 → HCF = 12
• LCM = 84 × 120 ÷ 12 = 84 × 10 = 840

Method 2: For Co-prime Numbers

If HCF(a, b) = 1, then LCM = a × b

• LCM(7, 13) = 91
• LCM(8, 15) = 120
• LCM(11, 17) = 187

Method 3: When One Divides the Other

LCM = the larger number.

• LCM(12, 36) = 36
• LCM(25, 100) = 100

Method 4: LCM of Three Numbers

Find LCM of first two, then find LCM of result with third.

Example: LCM(6, 10, 15)

• LCM(6, 10) = 30
• LCM(30, 15) = 30 (since 15 divides 30)
• Answer = 30

Part 4: HCF and LCM Problem Templates

Template 1: Largest Number That Divides Multiple Numbers Leaving Same Remainder

Formula: HCF of (differences between the numbers)

Worked Example:
Find the largest number that divides 77, 147, and 252 leaving the same remainder.

• Differences: 147−77 = 70, 252−147 = 105, 252−77 = 175
• HCF(70, 105): 105 = 1×70+35, 70 = 2×35 → HCF = 35
• HCF(35, 175): 175 = 5×35 → HCF = 35
• Answer = 35

Template 2: Largest Number That Divides Multiple Numbers Leaving Different Remainders

Formula: HCF of (each number minus its remainder)

Worked Example:
Find the largest number that divides 245 and 1029 leaving remainders 5 and 9 respectively.

• 245 − 5 = 240, 1029 − 9 = 1020
• HCF(240, 1020): 1020 = 4×240+60, 240 = 4×60+0 → HCF = 60
• Answer = 60

Template 3: Smallest Number Divisible by Multiple Numbers

Formula: LCM of all the numbers

Worked Example:
Smallest number divisible by 12, 18, and 30?

• LCM(12, 18) = 36
• LCM(36, 30): HCF(36, 30) = 6, LCM = 36×30÷6 = 180

Template 4: Bells/Events That Coincide

Formula: LCM gives the time when all events coincide

Worked Example:
Three bells ring at intervals of 12, 18, and 24 minutes. If they ring together at 8 AM, when do they next ring together?

• LCM(12, 18) = 36
• LCM(36, 24): HCF = 12, LCM = 36×24÷12 = 72
• They next ring together after 72 minutes = 9:12 AM

Part 5: Unit Digit Patterns

Unit digit questions appear frequently in TCS NQT and SSC exams. The key is knowing the cyclicity of each digit when raised to powers.

Cyclicity Table

How to Find Unit Digit of Any Power

1. Identify the unit digit of the base
2. Find the cyclicity period for that digit
3. Find (power mod period) — this gives position in the cycle
4. Look up that position in the cycle

Worked Example 1: Unit digit of 7⁴⁵

• Unit digit of base = 7, cycle = 7, 9, 3, 1 (period 4)
• 45 mod 4 = 1 → 1st position in cycle = 7

Worked Example 2: Unit digit of 3⁸⁶

• Cycle of 3: 3, 9, 7, 1 (period 4)
• 86 mod 4 = 2 → 2nd position = 9

Worked Example 3: Unit digit of 4¹²⁷

• Cycle of 4: 4, 6 (period 2)
• 127 mod 2 = 1 → 1st position = 4

Worked Example 4: Unit digit of 8⁹⁹ × 7⁴⁵

• 8⁹⁹: cycle 8, 4, 2, 6 (period 4), 99 mod 4 = 3 → 3rd position = 2
• 7⁴⁵: as above = 7
• Unit digit of product = unit digit of 2 × 7 = unit digit of 14 = 4

Part 6: Remainder Tricks

Trick 1: Remainder When Divided by 9 or 3 (Digital Root)

Remainder ÷ 9 = digital root of the number (sum of digits, reduced to single digit)

Remainder ÷ 3 = digital root mod 3

Examples:

• 4,729 ÷ 9: digit sum = 4+7+2+9 = 22 → 2+2 = 4 → remainder = 4
• 7,654 ÷ 3: digit sum = 22 → 22÷3 = remainder 1

Trick 2: Remainder When Divided by 11

Remainder = alternating digit sum (odd position sum − even position sum) mod 11

Example: 4,729 ÷ 11

• Odd positions: 4+2 = 6, Even positions: 7+9 = 16
• Difference: 6 − 16 = −10 → add 11 → remainder = 1

Trick 3: Pattern-Based Remainders for Powers

Fermat's Little Theorem (simplified for exams):
For prime p, aᵖ⁻¹ ÷ p leaves remainder 1 (when p does not divide a)

Practical application:

• 2¹⁰ ÷ 11: since 11 is prime, 2¹⁰ mod 11 = 1 → remainder = 1
• 3¹⁰ ÷ 11 = 1 (same logic)

Cyclicity of remainders — Worked Example: Remainder when 7³⁶ is divided by 5

• 7¹ mod 5 = 2
• 7² mod 5 = 4
• 7³ mod 5 = 3
• 7⁴ mod 5 = 1 → cycle repeats with period 4
• 36 mod 4 = 0 → 4th position in cycle = 1

Trick 4: Remainder of Sum/Product

Rule: Remainder of (a+b) ÷ n = (Remainder of a÷n + Remainder of b÷n) mod n

Rule: Remainder of (a×b) ÷ n = (Remainder of a÷n × Remainder of b÷n) mod n

Worked Example: Remainder when (253 × 347) is divided by 9

• 253 mod 9: digit sum = 10 → 1 → remainder = 1
• 347 mod 9: digit sum = 14 → 5 → remainder = 5
• Product remainder = 1 × 5 = 5

Part 7: Factor Counting and Sum of Factors

Counting Total Factors

Method:

• Express number as product of prime factors: N = pᵃ × qᵇ × rᶜ
• Total factors = (a+1)(b+1)(c+1)

Worked Examples:

Total factors of 360:

• 360 = 2³ × 3² × 5¹
• Total factors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24

Total factors of 1,260:

• 1260 = 2² × 3² × 5 × 7
• Total factors = (2+1)(2+1)(1+1)(1+1) = 3 × 3 × 2 × 2 = 36

Sum of All Factors

Formula: Sum = [(p^(a+1) − 1) ÷ (p−1)] × [(q^(b+1) − 1) ÷ (q−1)] × …

Worked Example: Sum of factors of 12

• 12 = 2² × 3¹
• Sum = [(2³ − 1) ÷ (2−1)] × [(3² − 1) ÷ (3−1)] = [7÷1] × [8÷2] = 7 × 4 = 28
• Verify: 1+2+3+4+6+12 = 28 ✓

Number of Even and Odd Factors

Odd factors: Set all powers of 2 to zero in the formula

• Odd factors of 360 = (2+1)(1+1) = 6 (from 3² × 5¹ only)

Even factors: Total factors − Odd factors

• Even factors of 360 = 24 − 6 = 18

Part 8: Consecutive Number Properties

These properties eliminate calculation entirely for certain question types.

Sum of Consecutive Numbers

• Sum of first n natural numbers = n(n+1) ÷ 2
• Sum of first n even numbers = n(n+1)
• Sum of first n odd numbers = n²

Quick applications:

• Sum of 1 to 50 = 50 × 51 ÷ 2 = 1,275
• Sum of first 20 odd numbers = 20² = 400
• Sum of first 15 even numbers = 15 × 16 = 240

Properties Used in Exam Problems

Property 1: Product of any n consecutive integers is divisible by n!

• Product of any 3 consecutive numbers is divisible by 6
• Product of any 4 consecutive numbers is divisible by 24

Property 2: Every even number is the sum of two prime numbers (Goldbach for small numbers):

• 8 = 3+5, 20 = 3+17, 28 = 5+23

Property 3: Square of any odd number leaves remainder 1 when divided by 8

• 7² = 49 → 49÷8 = remainder 1 ✓
• 11² = 121 → 121÷8 = remainder 1 ✓

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