LCM & HCF 

(Least Common Multiple & Highest Common Factor)

LCM and HCF are two fundamental properties of integers. HCF is the largest number that can completely
divide all given numbers, whereas LCM is the smallest number that is completely divisible by all given
numbers.

1. Highest Common Factor (HCF / GCD):
   The HCF of two or more numbers is the greatest number that divides each number without leaving any
   remainder.
   Example (for 12 and 18):
   Factors of 12 = 1, 2, 3, 4, 6, 12
   Factors of 18 = 1, 2, 3, 6, 9, 18
   Common factors = 1, 2, 3, 6
   HCF = 6
2. Least Common Multiple (LCM):
   The LCM of two or more numbers is the smallest non-zero number that is a multiple of each of the given
   numbers.
   Example (for 12 and 18):
   Multiples of 12 = 12, 24, 36, 48, 60, 72, …
   Multiples of 18 = 18, 36, 54, 72, 90, …
   Common multiples = 36, 72, 108, …
   LCM = 36
   Methods for Finding HCF and LCM:
3. Prime Factorization Method: First, express each number in terms of its prime factors.
   HCF: The product of the lowest powers of common prime factors.
   LCM: The product of the highest powers of all prime factors involved.
   Characteristics and Essential Formulas
   Formula 1 (Product of two numbers): For any two numbers 𝑎 and 𝑏, HCF(𝑎, 𝑏) × LCM(𝑎, 𝑏) = 𝑎 × 𝑏
   (Note: This rule applies only to two numbers, not for three or more numbers.)
   Formula 2 (HCF and LCM of fractions):
   HCF of fractions =HCF of numerators/LCM of denominators
   
   LCM of fractions =LCM of numerators/HCF of denominators
   
   Formula 3 (Co-prime Ratio Concept): If the HCF of two numbers 𝑎 and 𝑏 is 𝐻, then the numbers can be
   written as 𝑎 = 𝐻 × 𝑥 and 𝑏 = 𝐻 × 𝑦 (where 𝑥 and 𝑦 are co-prime, meaning HCF(𝑥, 𝑦) = 1). From this
   concept, LCM = 𝐻 × 𝑥 × 𝑦 = HCF × 𝑥 × 𝑦.
   Problem-Solving Formulas
   Type 1: HCF Problems (Group Division):
   Find the greatest number that divides 𝐴, 𝐵, 𝐶 leaving no remainder:HCF(𝐴, 𝐵, 𝐶).
   Find the greatest number that divides 𝐴, 𝐵, 𝐶 leaving remainders 𝑟1, 𝑟2, 𝑟3 respectively: HCF(𝐴 − 𝑟1,𝐵 − 𝑟2, 𝐶 −𝑟3).
   Find the greatest number that divides 𝐴, 𝐵, 𝐶 leaving the same remainder in each case: HCF(|𝐴 − 𝐵|, |𝐵 −𝐶|, |𝐶 − 𝐴|).
   
   Type 2: LCM Problems (Cycles / Minimum Quantity):
   Find the smallest number that is completely divisible by 𝐴, 𝐵, 𝐶: LCM(𝐴, 𝐵, 𝐶).
   Find the smallest number which when divided by 𝐴, 𝐵, 𝐶 leaves a remainder 𝑘 in each case: LCM(𝐴, 𝐵, 𝐶) + 𝑘.
   When bells toll together and then ring at intervals of 𝑇1, 𝑇2, 𝑇3, or traffic lights change: Next simultaneous
   occurrence time = LCM(𝑇1, 𝑇2, 𝑇3).
   
   Important Notes
   HCF × LCM = Product of two numbers.
   For any set of numbers, the HCF is always a factor of the LCM.
   For simultaneous events (such as bells ringing, traffic lights changing, running on a circular track), you need to
   find the LCM of the separate time intervals.

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PREVIOUS YEAR QUESTION

1. The HCF and LCM of two numbers are 12 and 180 respectively. If one number is 36, what is the other
   number?
   (A) 48
   (B) 60
   (C) 72
   (D) 90
   Answer: (B) 60
   Solution: According to the formula, 1st number × 2nd number = HCF × LCM.
   36 × 2nd number = 12 × 180 ⟹ 2nd number =12×180/
   36= 60.
2. The HCF of two numbers is 8. Which of the following can never be their LCM?
   (A) 24
   (B) 48
   (C) 56
   (D) 60
   Answer: (D) 60
   Solution: Let us divide the options by the HCF (8):
   24 ÷ 8 = 3 (Divisible)
   48 ÷ 8 = 6 (Divisible)
   56 ÷ 8 = 7 (Divisible)
   60 ÷ 8 = 7.5 (Not completely divisible, so it cannot be the LCM)
3. What is the greatest number which, when it divides 1356, 1868, and 2764, leaves a remainder of 12 in each
   case?
   (A) 64
   (B) 124
   (C) 156
   (D) 260
   Answer: (C) 156
   Solution: Subtract 12 from each number:
   1356 − 12 = 1344, 1868 − 12 = 1856, 2764 − 12 = 2752.
   Now we find the HCF of the mutual differences of these numbers:
   1856 − 1344 = 512, 2752 − 1856 = 896.
   HCF(512,896) = 156.
4. Three bells ring at intervals of 9, 12, and 15 minutes respectively. If they ring together now, after how many hours will they ring together again?
   (A) 3 hours
   (B) 6 hours
   (C) 9 hours
   (D) 12 hours
   Answer: (A) 3 hours
   Solution: Find the LCM of 9, 12, and 15:
   9 = 3×3,
   12 = 2×2 × 3,
   15 = 3 × 5.
   LCM = 2×2 ×3×3× 5=4 × 9 × 5 = 180 minutes.
   Converting minutes into hours: 180 ÷ 60 = 3 hours.
5. Which is the smallest square number that is completely divisible by 3, 4, 5, 6, and 8?
   (A) 900
   (B) 1200
   (C) 2500
   (D) 3600
   Answer: (D) 3600
   Solution: LCM(3,4,5,6,8) = 120.
   Prime factorization of 120 = 2×3 × 3 × 5.
   To make it a perfect square, multiply by the unmatched prime factors (2,3,5), which gives 2 × 3 × 5 = 30.
   Required square number = 120 × 30 = 3600.
6. The ratio of two numbers is 3:4. If their HCF is 4, what is their LCM?
   (A) 12
   (B) 16
   (C) 24
   (D) 48
   Answer: (D) 48
   Solution: Let the numbers be 3𝑥 and 4𝑥, where 𝑥 = HCF = 4.
   LCM = HCF × Product of ratio terms = 4 × 3 × 4 = 48.
7. Find the HCF of 2/3,8/9,16/81, and 10/27.
   (A) 2/81
   (B) 16/3
   (C) 2/3
   (D) 80/81
   Answer: (A) 2/81
   Solution: HCF of fractions =HCF of numerators/LCM of denominators.
   HCF(2,8,16,10) = 2.
   LCM(3,9,81,27) = 81.
   Required HCF =2/81.
8. Find the smallest number which when divided by 12, 16, 18, and 21 leaves a remainder of 7 in each case.
   (A) 1008
   (B) 1015
   (C) 1022
   (D) 1032
   Answer: (B) 1015
   Solution: LCM(12,16,18,21) = 1008.
   Add the remainder: Required number = 1008 + 7 = 1015.
9. What is the largest 4-digit number that is completely divisible by 12, 15, 18, and 27?
   (A) 9720
   (B) 9930
   (C) 9860
   (D) 9960
   Answer: (A) 9720
   Solution: LCM(12,15,18,27) = 540.
   The largest 4-digit number is 9999.
   Dividing 9999 by 540 leaves a remainder of 279.
   Required number = 9999 − 279 = 9720.
   
10. The sum of two numbers is 384 and their HCF is 48. How many such pairs of numbers are possible?
    (A) 1
    (B) 2
    (C) 3
    (D) 4
    Answer: (B) 2
    Solution: Let the numbers be 48𝑎 and 48𝑏 (where 𝑎 and 𝑏 are co-prime).
    48(𝑎 + 𝑏) = 384 ⟹ 𝑎 + 𝑏 = 8.
    Co-prime pairs whose sum is 8 are: (1,7) and (3,5). Thus, only 2 pairs are possible.
11. The LCM of two numbers is 12 times their HCF. The sum of HCF and LCM is 403. If one number is 93,
    what is the other number?
    (A) 124
    (B) 134
    (C) 128
    (D) 138
    Answer: (A) 124
    Solution: LCM + HCF = 12 × HCF + HCF = 13 × HCF = 403 ⟹ HCF = 31.
    LCM = 12 × 31 = 372.
    Other number =HCF×LCM
    Given number =(31×372)/93= 124.
12. Traffic lights at three different road crossings change every 48 seconds, 72 seconds, and 108 seconds
    respectively. If they change together at 8:20:00 AM, when will they change together again?
    (A) 8:27:12 AM
    (B) 8:25:20 AM
    (C) 8:24:12 AM
    (D) 8:28:00 AM
    Answer: (A) 8:27:12 AM
    Solution: LCM(48,72,108) = 432 seconds = 7 minutes 12 seconds.
    Exact time = 8:20:00 + 0:07:12 = 8:27:12.
13. Find the smallest number which when divided by 20, 25, 35, and 40 leaves remainders of 14, 19, 29, and 34
    respectively?
    (A) 1394
    (B) 1404
    (C) 1396
    (D) 1400
    Answer: (A) 1394
    Solution: Common difference between divisor and remainder = 20 − 14 = 6. LCM of 20, 25, 35, and 40 =
14. What is the LCM of 1/3,2/9,5/6, and 4/27?
    (A) 1/54
    (B) 10/3
    (C) 20/3
    (D) 20/27
    Answer: (C) 20/3
    Solution: LCM of fractions = LCM of numerators/HCF of denominators =LCM(1,2,5,4)/HCF(3,9,6,27)= 20/3.
15. What is the greatest number that divides 62, 132, and 237 leaving the same remainder in each case?
    (A) 15
    (B) 25
    (C) 35
    (D) 45
    Answer: (C) 35
    Solution: Take the HCF of the mutual differences: |132 − 62| = 70, |237 − 132| = 105, |237 − 62| =HCF(70,105,175) = 35.