HCF and LCM – VCA

1. Basic Concepts

HCF (Highest Common Factor): Largest number that divides all given numbers exactly. LCM (Lowest Common Multiple): Smallest number that is divisible by all given numbers.

Key Formula: HCF × LCM = Product of two numbers (for two numbers only)

2. Methods to Find HCF

Method 1: Prime Factorization

Q1. Find HCF of 48, 72 Solution:

48 = 2⁴ × 3¹
72 = 2³ × 3²
HCF = 2³ × 3¹ = 8 × 3 = 24

Q2. Find HCF of 60, 84, 108 Solution:

60 = 2² × 3¹ × 5¹
84 = 2² × 3¹ × 7¹
108 = 2² × 3³
HCF = 2² × 3¹ = 4 × 3 = 12

Q3. Find HCF of 144, 180, 240 Solution:

144 = 2⁴ × 3²
180 = 2² × 3² × 5¹
240 = 2⁴ × 3¹ × 5¹
HCF = 2² × 3¹ = 4 × 3 = 12

Q4. Find HCF of 126, 162, 198 Solution:

126 = 2¹ × 3² × 7¹
162 = 2¹ × 3⁴
198 = 2¹ × 3² × 11¹
HCF = 2¹ × 3² = 2 × 9 = 18

Q5. Find HCF of 75, 125, 200 Solution:

75 = 3¹ × 5²
125 = 5³
200 = 2³ × 5²
HCF = 5² = 25

Method 2: Euclidean Algorithm (Division Method)

Q6. Find HCF of 56, 98 using Euclidean method Solution:

98 = 56 × 1 + 42
56 = 42 × 1 + 14
42 = 14 × 3 + 0
HCF = 14

Q7. Find HCF of 144, 192 using division method Solution:

192 = 144 × 1 + 48
144 = 48 × 3 + 0
HCF = 48

Q8. Find HCF of 210, 315 using Euclidean algorithm Solution:

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

Q9. Find HCF of 168, 252 Solution:

252 = 168 × 1 + 84
168 = 84 × 2 + 0
HCF = 84

Q10. Find HCF of 391, 667 Solution:

667 = 391 × 1 + 276
391 = 276 × 1 + 115
276 = 115 × 2 + 46
115 = 46 × 2 + 23
46 = 23 × 2 + 0
HCF = 23

Method 3: Factorization Method

Q11. Find HCF of 36, 54 by listing factors Solution:

Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54
Common factors: 1, 2, 3, 6, 9, 18
HCF = 18

Q12. Find HCF of 42, 70 by factorization Solution:

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42
Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70
Common factors: 1, 2, 7, 14
HCF = 14

3. Methods to Find LCM

Method 1: Prime Factorization

Q13. Find LCM of 48, 72 Solution:

48 = 2⁴ × 3¹
72 = 2³ × 3²
LCM = 2⁴ × 3² = 16 × 9 = 144

Q14. Find LCM of 12, 18, 24 Solution:

12 = 2² × 3¹
18 = 2¹ × 3²
24 = 2³ × 3¹
LCM = 2³ × 3² = 8 × 9 = 72

Q15. Find LCM of 15, 25, 35 Solution:

15 = 3¹ × 5¹
25 = 5²
35 = 5¹ × 7¹
LCM = 3¹ × 5² × 7¹ = 3 × 25 × 7 = 525

Q16. Find LCM of 20, 30, 40 Solution:

20 = 2² × 5¹
30 = 2¹ × 3¹ × 5¹
40 = 2³ × 5¹
LCM = 2³ × 3¹ × 5¹ = 8 × 3 × 5 = 120

Q17. Find LCM of 14, 21, 28 Solution:

14 = 2¹ × 7¹
21 = 3¹ × 7¹
28 = 2² × 7¹
LCM = 2² × 3¹ × 7¹ = 4 × 3 × 7 = 84

Method 2: Division Method

Q18. Find LCM of 36, 48, 60 using division method Solution:

2 | 36, 48, 60
2 | 18, 24, 30
2 |  9, 12, 15
3 |  9, 12, 15
2 |  3,  4,  5
2 |  3,  2,  5
5 |  3,  1,  5
3 |  3,  1,  1
  |  1,  1,  1

LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Q19. Find LCM of 16, 24, 32 Solution:

2 | 16, 24, 32
2 |  8, 12, 16
2 |  4,  6,  8
2 |  2,  3,  4
2 |  1,  3,  2
3 |  1,  3,  1
  |  1,  1,  1

LCM = 2⁵ × 3 = 32 × 3 = 96

4. Using HCF × LCM Formula

Q20. Two numbers are 72 and 96. Find their HCF and LCM. Solution:

HCF(72,96): 72 = 2³×3², 96 = 2⁵×3¹, HCF = 2³×3¹ = 24
LCM using formula: LCM = (72×96)÷24 = 6912÷24 = 288

Q21. Product of two numbers is 4800 and their HCF is 16. If one number is 80, find the other and their LCM. Solution:

Let other number = x
80 × x = 4800, so x = 60
LCM = (80×60)÷16 = 4800÷16 = 300

Q22. LCM of two numbers is 180 and HCF is 12. If one number is 36, find the other. Solution:

HCF × LCM = Product of numbers
12 × 180 = 36 × other number
Other number = 2160÷36 = 60

Q23. Two numbers are in ratio 3:4. Their LCM is 84. Find the numbers and their HCF. Solution:

Let numbers be 3x and 4x
HCF = x (since 3 and 4 are coprime)
LCM = 3x × 4x ÷ x = 12x
12x = 84, so x = 7
Numbers are 21 and 28, HCF = 7

5. Word Problems – Time and Distance

Q24. Three bells ring at intervals of 9, 12, 15 minutes. If they ring together at 8:00 AM, when will they ring together again? Solution:

LCM(9,12,15) = LCM(3²,2²×3,3×5) = 2²×3²×5 = 180 minutes = 3 hours
Next time: 8:00 AM + 3 hours = 11:00 AM

Q25. Two wheels make 120 and 180 revolutions per minute. After how many seconds will they be together at the starting point? Solution:

LCM(120,180) = LCM(2³×3×5, 2²×3²×5) = 2³×3²×5 = 360
Time = 360÷360 = 1 minute = 60 seconds

Q26. Four buses start together and run at speeds to complete rounds in 6, 8, 12, 18 minutes respectively. After how much time will they meet again at starting point? Solution:

LCM(6,8,12,18) = LCM(2×3, 2³, 2²×3, 2×3²) = 2³×3² = 72 minutes = 1 hour 12 minutes

Q27. A traffic light changes every 48 seconds, another every 72 seconds. If both change together at 2:00 PM, when will they change together again? Solution:

LCM(48,72) = LCM(2⁴×3, 2³×3²) = 2⁴×3² = 144 seconds = 2 minutes 24 seconds
Next time: 2:02:24 PM

Q28. Two runners complete a circular track in 8 and 12 minutes respectively. If they start together, after how much time will they meet again at starting point? Solution:

LCM(8,12) = LCM(2³, 2²×3) = 2³×3 = 24 minutes

6. Word Problems – Measurement and Division

Q29. Find the largest number that divides 245, 1029, 1458 leaving remainders 5, 9, 8 respectively. Solution:

Numbers after removing remainders: 240, 1020, 1450
HCF(240,1020,1450) = 10

Q30. Three pieces of timber 42m, 49m, 63m long have to be divided into planks of equal length. What is the greatest possible length of each plank? Solution:

HCF(42,49,63) = HCF(2×3×7, 7², 3²×7) = 7
Greatest length = 7m

Q31. Find the greatest number that will divide 43, 91, 183 so as to leave the same remainder in each case. Solution:

Differences: 91-43=48, 183-91=92, 183-43=140
HCF(48,92,140) = 4

Q32. The HCF of two numbers is 16 and their product is 3072. Find their LCM. Solution:

LCM = Product ÷ HCF = 3072 ÷ 16 = 192

Q33. A rectangular hall 18m long and 12m wide is to be paved with square tiles of equal size. Find the largest size of each tile. Solution:

HCF(18,12) = 6
Largest tile size = 6m × 6m

Q34. Find the smallest number which leaves remainders 8, 12, 20 when divided by 12, 16, 24 respectively. Solution:

Required remainders are 4 less than divisors in each case
LCM(12,16,24) = 48
Required number = 48 – 4 = 44

7. Co-prime Numbers and HCF

Q35. Check if 15 and 28 are co-prime. Solution:

HCF(15,28) = HCF(3×5, 2²×7) = 1
Yes, they are co-prime

Q36. Find pairs of co-prime numbers between 10 and 20. Solution:

Check each pair: (11,12), (11,13), (11,14), (11,15), (11,16), (11,17), (11,18), (11,19)
(13,14), (13,15), (13,16), (13,17), (13,18), (13,19)
And so on… Many pairs exist.

Q37. If HCF of two numbers is 1, what can you say about their LCM? Solution:

If HCF = 1, then LCM = Product of the numbers

Q38. Two numbers are co-prime. Their product is 1001. Find the numbers. Solution:

1001 = 7 × 11 × 13
Possible pairs: (1,1001), (7,143), (11,91), (13,77)

8. Advanced Problems

Q39. Find HCF and LCM of 2⁴×3²×5, 2³×3×5², 2²×3³×5³ Solution:

HCF = 2²×3¹×5¹ = 4×3×5 = 60
LCM = 2⁴×3³×5³ = 16×27×125 = 54000

Q40. Three numbers are in ratio 1:2:3. Their LCM is 60. Find the numbers. Solution:

Let numbers be x, 2x, 3x
LCM = 6x = 60, so x = 10
Numbers are 10, 20, 30

Q41. Four numbers are in ratio 2:3:4:5. Their LCM is 240. Find HCF. Solution:

Let numbers be 2x, 3x, 4x, 5x
LCM = 60x = 240, so x = 4
Numbers are 8, 12, 16, 20
HCF = 4

Q42. Find the smallest number that leaves remainder 1 when divided by 2, 3, 4, 5, 6. Solution:

LCM(2,3,4,5,6) = 60
Required number = 60 + 1 = 61

Q43. Two numbers differ by 6. Their product is 1008 and HCF is 6. Find the numbers. Solution:

Let numbers be 6a and 6b where HCF(a,b) = 1
6b – 6a = 6, so b – a = 1
6a × 6b = 1008, so ab = 28
Since b = a + 1: a(a+1) = 28, so a = 7, b = 8
Numbers are 42 and 48

Q44. Find LCM of first 10 natural numbers. Solution:

Numbers: 1,2,3,4,5,6,7,8,9,10
Prime factorization method: 2³×3²×5×7 = 2520

Q45. The LCM of two numbers is 14 times their HCF. Sum of LCM and HCF is 600. Find HCF and LCM. Solution:

Let HCF = h, then LCM = 14h
h + 14h = 600
15h = 600, so h = 40
HCF = 40, LCM = 560

9. Fraction Problems

Q46. Reduce 144/180 to lowest terms. Solution:

HCF(144,180) = 36
Reduced fraction = 144÷36 / 180÷36 = 4/5

Q47. Find LCM of denominators: 1/6, 1/8, 1/12, 1/15 Solution:

LCM(6,8,12,15) = LCM(2×3, 2³, 2²×3, 3×5) = 2³×3×5 = 120

Q48. Simplify: 126/189 Solution:

HCF(126,189) = 63
Simplified: 126÷63 / 189÷63 = 2/3

Q49. Find the smallest number that can be expressed as sum of 2, 3, 4, 5, 6 equal parts. Solution:

LCM(2,3,4,5,6) = 60

10. Multiple Number Problems

Q50. Find HCF of 84, 112, 196 Solution:

84 = 2² × 3 × 7
112 = 2⁴ × 7
196 = 2² × 7²
HCF = 2² × 7 = 28

Q51. Find LCM of 45, 60, 75 Solution:

45 = 3² × 5
60 = 2² × 3 × 5
75 = 3 × 5²
LCM = 2² × 3² × 5² = 900

Q52. Find HCF and LCM of 36, 54, 72 Solution:

36 = 2² × 3²
54 = 2 × 3³
72 = 2³ × 3²
HCF = 2 × 3² = 18
LCM = 2³ × 3³ = 216

Q53. Find the smallest number divisible by 8, 12, 15, 20 Solution:

LCM(8,12,15,20) = LCM(2³, 2²×3, 3×5, 2²×5) = 2³×3×5 = 120

Q54. Find HCF of 168, 180, 264 Solution:

168 = 2³ × 3 × 7
180 = 2² × 3² × 5
264 = 2³ × 3 × 11
HCF = 2² × 3 = 12

11. Pattern and Sequence Problems

Q55. If HCF(a,b) = 12 and LCM(a,b) = 180, find possible values of a and b. Solution:

a×b = 12×180 = 2160
Let a = 12m, b = 12n where HCF(m,n) = 1
12m × 12n = 2160, so mn = 15
Possible pairs (m,n): (1,15), (3,5), (5,3), (15,1)
So (a,b): (12,180), (36,60), (60,36), (180,12)

Q56. Three numbers are such that their pairwise HCFs are 6, 8, 12. Find the smallest possible values. Solution:

Let numbers be 6a, 8b, 12c where appropriate conditions hold
After analysis: smallest values are 24, 48, 72

Q57. Find numbers whose HCF is 15 and LCM is 180. Solution:

Let numbers be 15a and 15b where HCF(a,b) = 1
LCM = 15ab = 180, so ab = 12
Possible pairs: (1,12), (3,4), (4,3), (12,1)
Numbers: (15,180), (45,60), (60,45), (180,15)

12. Time-Based Problems

Q58. A clock strikes once at 1 o’clock, twice at 2 o’clock, and so on. How many times will it strike in 24 hours? Solution:

Not directly HCF/LCM but: 1+2+3+…+12 (twice) = 2×78 = 156

Q59. Three alarm clocks ring at intervals of 20, 24, 30 minutes. If they ring together at 6:00 AM, find next 3 times they ring together. Solution:

LCM(20,24,30) = 120 minutes = 2 hours
Times: 8:00 AM, 10:00 AM, 12:00 PM

Q60. Two pendulums complete oscillations in 3 and 4 seconds. After how much time will they be in phase again? Solution:

LCM(3,4) = 12 seconds

13. Division and Remainder Problems

Q61. Find the greatest number that divides 650 and 1170 leaving remainders 5 and 7 respectively. Solution:

Numbers after removing remainders: 645, 1163
HCF(645,1163) = 1 (they are coprime)
Answer: 1

Q62. What is the largest number that divides 280, 1245, 2008 leaving same remainder? Solution:

Find differences: 1245-280=965, 2008-1245=763, 2008-280=1728
HCF(965,763,1728) = 1

Q63. Find the smallest number which when divided by 6, 8, 12 leaves remainder 3 in each case. Solution:

LCM(6,8,12) = 24
Required number = 24k + 3, smallest = 27

14. Geometric Problems

Q64. A rectangular field 180m by 105m is to be divided into square plots of equal size. Find the largest possible size of each plot. Solution:

HCF(180,105) = 15
Largest square plot = 15m × 15m

Q65. Three rods of lengths 96cm, 144cm, 192cm are to be cut into pieces of equal length. Find the greatest possible length of each piece. Solution:

HCF(96,144,192) = 48
Greatest length = 48cm

Q66. A room 15m long, 12m wide, 9m high is to be divided into cubical blocks of equal size. Find the side of largest cube. Solution:

HCF(15,12,9) = 3
Side of largest cube = 3m

15. Advanced Applications

Q67. In how many ways can 240 be expressed as LCM of two numbers? Solution:

Find all factor pairs and check which give LCM = 240
Requires detailed analysis of all factor combinations

Q68. If LCM of three numbers is 9600 and their HCF is 8, find the maximum possible product of the three numbers. Solution:

Complex problem requiring optimization techniques

Q69. Two numbers are such that their HCF is 16 and LCM is 480. How many such pairs exist? Solution:

Product = 16 × 480 = 7680
Let numbers be 16a, 16b where HCF(a,b) = 1
ab = 30 = 2×3×5
Number of ways = 2³ = 8, so 4 unordered pairs

Q70. Find the smallest positive integer that leaves remainder 2 when divided by 6, remainder 3 when divided by 7, remainder 4 when divided by 8. Solution:

Using Chinese Remainder Theorem or systematic checking
Answer: 164

16. Speed Problems

Q71. Two cars start from same point in same direction. One travels at 40 km/h, other at 60 km/h. After how much time will the faster car be exactly one lap ahead on a circular track of 12 km? Solution:

Relative speed = 60-40 = 20 km/h
Time = 12/20 = 0.6 hours = 36 minutes

Q72. Three cyclists start together and cycle at speeds to complete one round in 4, 6, 8 minutes. When will they meet again? Solution:

LCM(4,6,8) = 24 minutes

17. Mixed Practice Problems

Q73. If HCF(a,b) = 18 and a+b = 126, find possible values of a and b. Solution:

Let a = 18m, b = 18n where HCF(m,n) = 1
18m + 18n = 126, so m + n = 7
Possible pairs: (1,6), (2,5), (3,4) and reverse
Values: (18,108), (36,90), (54,72) and reverse

Q74. The ratio of two numbers is 3:4. If their LCM is 84, find their HCF. Solution:

Let numbers be 3x, 4x
HCF = x, LCM = 12x = 84
x = 7, so HCF = 7

Q75. Find the smallest number that leaves remainder 1 when divided by 2, remainder 2 when divided by 3, remainder 3 when divided by 4. Solution:

Pattern: remainder is one less than divisor
LCM(2,3,4) = 12
Required number = 12 – 1 = 11

Q76. Two numbers are in ratio 4:5. Their LCM is 180. Find the numbers. Solution:

Let numbers be 4x, 5x
LCM = 20x = 180, so x = 9
Numbers are 36, 45

Q77. Find HCF of 2520, 3780, 4200 Solution:

2520 = 2³ × 3² × 5 × 7
3780 = 2² × 3³ × 5 × 7
4200 = 2³ × 3 × 5² × 7
HCF = 2² × 3 × 5 × 7 = 420

Q78. The LCM of two numbers is 165 and their ratio is 3:11. Find the numbers. Solution:

Let numbers be 3x, 11x
LCM = 33x = 165, so x = 5
Numbers are 15, 55

Q79. Find the smallest 4-digit number divisible by 18, 24, 32. Solution:

LCM(18,24,32) = 288
Smallest 4-digit = 1000
1000 ÷ 288 = 3.47, so 4 × 288 = 1152

Q80. Three taps can fill a tank in 12, 15, 20 minutes respectively. If opened together, find when the tank will be full. Solution:

Rates: 1/12, 1/15, 1/20 per minute
Combined rate = 5+4+3/60 = 12/60 = 1/5
Time = 5 minutes

Quick Reference Formulas:

For two numbers: HCF × LCM = Product of numbers
Prime factorization: HCF = product of lowest powers, LCM = product of highest powers
For coprime numbers: HCF = 1, LCM = Product
Multiple numbers: Use prime factorization method
Euclidean algorithm: Divide larger by smaller repeatedly until remainder is 0
Word problems: Identify whether to find HCF (division/measurement) or LCM (meeting/timing)
Remainder problems: Subtract remainder first, then find HCF
Ratio problems: If numbers are in ratio a:b, let them be ax and bx

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