Grasping the Concept of LCM becomes easier when you practice with a range of questions. Here, we provide a variety of LCM questions, all complete with comprehensive solutions and explanations, to aid your understanding. To delve deeper into the concept of LCM, visit What is LCM?

The LCM, or Lowest Common Multiple, of two or more numbers is the smallest common multiple of the given numbers. For instance, the LCM of 3, 5 and 7 is 105, which is the smallest number that all of 3, 5 and 7 can divide.

• Finding LCM by listing all multiples.

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Consider the LCM of 6 and 9, the multiples of 6 are 6, 12, 18, 24… and the multiples of 9 are 9, 18, 27, 36 … The smallest common multiple is 18. However, this method becomes less efficient with larger numbers.

• Finding LCM by prime factorisation.

Let's find the LCM of 14 and 18

Prime factorisation of 14 = 2 × 7

Prime factorisation of 18 = 2 × 3 × 3 = 2 × 3 2

We take the highest power of the prime factors of both numbers, that is, 2 × 3 2 × 7 = 126

Important Formulas for Finding LCM

1. Prime Factorization Method:
   Break each number into its prime factors (these are the basic numbers that multiply to give the original number).
   Then, take the highest power of each prime factor from both numbers and multiply them together. This gives the LCM.
    

2. Using Greatest Common Divisor (GCD):
   Another way is to multiply the two numbers together and then divide by their GCD (the largest number that divides both without leaving a remainder).
   The formula is:
   LCM(a, b) = (|a × b|) ÷ GCD(a, b)

1. If HCF(200, 500) = 100, find LCM(200, 500).

Solution: We have LCM of two numbers = (Product of two numbers)/ their HCF

= (200 × 500)/100 = 1000.

Hence, LCM(200, 500) = 1000

The product of the LCM (Least Common Multiple) and HCF (Highest Common Factor) of two numbers is always equal to the product of those two numbers.

In other words, if you have two numbers, say a and b, then:

LCM(a, b) × HCF(a, b) = a × b

Note: This rule works only when you are dealing with two numbers.

2. Two runners are running around a circular track. Runner 1 takes 20 minutes to complete one lap, while Runner 2 completes the lap in 25 minutes. If they both start simultaneously and run in the same direction, after what duration will they meet at the starting point?

Solution: Time taken by the runners to meet again = LCM(20, 25)

Now 20 = 2 2 × 5 and 25 = 5 2

Therefore, LCM(20, 25) = 2 2 × 5 2 = 100

Hence, they will meet at the starting point after 100 minutes.

3. Is it feasible to have two numbers whose HCF is 24 and LCM is 720?

Solution: Since the HCF always divides the LCM, we see that 720 is divisible by 24, 720/24 = 30.

Thus, it is possible to have two numbers.

LCM Problems Involving Remainders

1. When a number leaves the same remainder
   If a number is divided by several numbers (like n₁, n₂, and so on) and it leaves the same remainder R each time, then:
   Number = LCM of (n₁, n₂, …) + R
    
2. When a number leaves different remainders
   If a number is divided by several numbers (n₁, n₂, …) and leaves different remainders (R₁, R₂, …) for each, but the difference between each divisor and its remainder is the same, that is:
   n₁ – R₁ = n₂ – R₂ = … = some number R,
   then the number can be found by:
   Number = LCM of (n₁, n₂, …) – R

4. Find the smallest number divided by 30 and 35, leaving the remainder 10 and 15, respectively.

Solution: Since 30 – 10 = 20 and 35 – 15 = 20

So we need LCM{(30, 35) – 20} = 210 – 20 = 190

Thus, 190 is the required number.

5. Find the smallest number, which, when divided by 40, 60 and 90, leaves the same remainder of 10, respectively.

Solution:Find LCM of 40, 60, and 90.

• 40 = 2³ × 5
• 60 = 2² × 3 × 5
• 90 = 2 × 3² × 5

• LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360
• Since the remainder is 10, the required number = LCM + remainder = 360 + 10 = 370..

6. Two timers ring their alarms at regular intervals of 80 seconds and 60 seconds. If they beep together at 8 am, at what time will they beep again for the first time?

Solution: We find the LCM of 80 and 60.

Prime factorisation of 80 and 60,

80 = 2 × 2 × 2 × 2 × 5

60 = 2 × 2 × 3 × 5

Therefore, the LCM of 80 and 60 = 2 4 × 3 × 5 = 240

240 seconds = 240/60 min = 4 min

Hence, the timers will beep again for the first time at 8:04 am.

7. There are 60 students in section A and 65 students in section B of a class in a school. Find the minimum number of books required for their class library so that they can be distributed equally among the students of section A or section B.

Solution: Clearly, the number of books to be equally distributed should be a multiple of 60 and of 65. Thus, we have to find the LCM of 60 and 65.

Now, 60 = 2 × 2 × 3 × 5

65 = 5 × 13

LCM (60, 65) = 2 2 × 3 × 5 × 13 = 780.

Hence, at least 780 books are required in the library.

Also Read:

• Euclid’s Division Lemma
• Real Numbers
• Rational and Irrational Numbers
• Polynomials

LCM of Polynomials

To find the LCM (Least Common Multiple) of polynomials, follow these easy steps:

1. Break down each polynomial into smaller parts called irreducible factors (factors that can’t be broken down further).
    
2. Look at all these factors from every polynomial and pick each factor only once. If a factor appears with different powers (exponents), choose the highest power.
    
3. Multiply all these factors together with their highest powers to get the LCM of the polynomials.

8. Find the LCM of 120(x – 2)(x + 2) 2 (x + 4) 3 and 80(x 2 – 2)(x + 4)(x + 3) 2 .

Solution: Let f(x) = 120(x – 2)(x + 2) 2 (x + 4) 3

And g(x) = 80(x 2 – 2)(x + 4)(x + 3) 2

Now, factorising the polynomials into irreducible factors.

f(x) = 2 × 2 × 2 × 3 × 5 × (x – 2)(x + 2) 2 (x + 4) 3

g(x) = 2 × 2 × 2 × 2 × 5 × (x + 2)(x – 2)(x + 4)(x + 3) 2

Taking all the factors raised to their highest exponents: 2 4 , 3, 5, (x – 2), (x + 2) 2 , (x + 4) 3 , (x + 3) 2

⇒ The LCM of the given polynomials = 240(x – 2)(x + 2) 2 (x + 3) 2 (x + 4) 3 .

Types of LCM:

• LCM for Whole Numbers (Integers):
  Find the LCM just like usual, but remember to consider if the numbers are positive or negative.
• LCM for Fractions:
  Use this formula:
  LCM = (LCM of the numerators) ÷ (HCF of the denominators)
• LCM for Decimals:
  First, change the decimals into whole numbers by multiplying by powers of 10.
  Then find the LCM of those whole numbers.
  Finally, put the decimal point back in the answer where it belongs.

9. Find the LCM of ⅔, 2/3 and 5/6.

Solution:

• LCM of numerators (2, 2, 5) = 10
   
• HCF of denominators (3, 3, 6) = 3
   
• LCM of fractions = (LCM of numerators) ÷ (HCF of denominators) = 10 / 3 ≈ 3.33 or 10/3.

10. Find the LCM of 22.5, 4.5 and 0.75.

Solution: Converting the decimals into integers,

22.5 = 22.5 × 100 = 2250

4.5 = 4.5 × 100 = 450

0.75 × 100 = 75

Now, 2250 = 2 × 3 × 3 × 5 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

75 = 3 × 5 × 5

LCM (225, 450, 75) = 2 × 3 2 × 5 3 = 2250

Place a decimal point two places from right

Then, LCM(22.5, 4.5, 0.75) = 22.5.

11. Find the smallest number, which, when reduced by 9, is divisible by 13, 17, 19, 23 and 29.

Solution:

• Find LCM of 13, 17, 19, 23, and 29:
  • 13, 17, 19, 23, 29 are prime numbers.
  • LCM = 13 × 17 × 19 × 23 × 29 = 122,521
• Let the number be x. Then:
  x - 9 = multiple of LCM → x = LCM + 9 = 122,521 + 9 = 122,530.

12. Find the largest four-digit number, which is divided by 5, 8 and 12, leaving a remainder of 4, respectively.

Solution: Largest 4-digit number = 9999

LCM (5, 8, 12) = 120

Now 9999 = 83 × 120 + 99

Thus, the 4-digit number divisible by 5, 8 and 12 = 9999 – 99 = 9900.

Since the number we have to find leaves a remainder of four when divided by 5, 8 and 12,

∴ 9900 + 4 = 9904 is the required number.

13.Find the LCM of 9, 6, and 15.

Solution:

1. Prime factorization:

• 9 = 3²
• 6 = 2 × 3
• 15 = 3 × 5

1. Take the highest power of each prime:

• For 2: 2¹
• For 3: 3²
• For 5: 5¹

1. Calculate LCM:

LCM = 2 × 3² × 5 = 2 × 9 × 5 = 90

Answer: The LCM of 9, 6, and 15 is 90.

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