Finding Common Ground | FIO

Question 6

Find the LCM of the following numbers:

(a) 30, 72

(b) 36, 54

(c) 105, 195, 65

(d) 222, 370

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Solution
Understand the Question
  • To find the Least Common Multiple (LCM) of a set of numbers using the prime factorisation method:
    1. Find the prime factors of each number and express them in exponential form.
    2. Identify all unique prime factors across the numbers.
    3. Take the highest power of each prime factor and multiply them together.

(a) 30, 72

Step 1 · Prime Factorisation of 30 and 72

For 3030:

30÷2=1515÷3=55÷5=1\begin{aligned} 30 \div 2 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \end{aligned}

30=2×3×530 = 2 \times 3 \times 5

For 7272:

72÷2=3636÷2=1818÷2=99÷3=33÷3=1\begin{aligned} 72 \div 2 &= 36 \\ 36 \div 2 &= 18 \\ 18 \div 2 &= 9 \\ 9 \div 3 &= 3 \\ 3 \div 3 &= 1 \end{aligned}

72=2×2×2×3×3=23×3272 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2

Step 2 · Calculate LCM

Taking the highest power of each prime factor (232^3, 323^2, 515^1):

LCM(30,72)=23×32×51=8×9×5=72×5=360\begin{aligned} \operatorname{LCM}(30, 72) &= 2^3 \times 3^2 \times 5^1 \\ &= 8 \times 9 \times 5 \\ &= 72 \times 5 \\ &= 360 \end{aligned}
Answer

(a) 360

(b) 36, 54

Step 1 · Prime Factorisation of 36 and 54

For 3636:

36÷2=1818÷2=99÷3=33÷3=1\begin{aligned} 36 \div 2 &= 18 \\ 18 \div 2 &= 9 \\ 9 \div 3 &= 3 \\ 3 \div 3 &= 1 \end{aligned}

36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2

For 5454:

54÷2=2727÷3=99÷3=33÷3=1\begin{aligned} 54 \div 2 &= 27 \\ 27 \div 3 &= 9 \\ 9 \div 3 &= 3 \\ 3 \div 3 &= 1 \end{aligned}

54=2×3×3×3=21×3354 = 2 \times 3 \times 3 \times 3 = 2^1 \times 3^3

Step 2 · Calculate LCM

Taking the highest power of each prime factor (222^2, 333^3):

LCM(36,54)=22×33=4×27=108\begin{aligned} \operatorname{LCM}(36, 54) &= 2^2 \times 3^3 \\ &= 4 \times 27 \\ &= 108 \end{aligned}
Answer

(b) 108

(c) 105, 195, 65

Step 1 · Prime Factorisation of 105, 195, and 65

For 105105:

105÷3=3535÷5=77÷7=1\begin{aligned} 105 \div 3 &= 35 \\ 35 \div 5 &= 7 \\ 7 \div 7 &= 1 \end{aligned}

105=3×5×7105 = 3 \times 5 \times 7

For 195195:

195÷3=6565÷5=1313÷13=1\begin{aligned} 195 \div 3 &= 65 \\ 65 \div 5 &= 13 \\ 13 \div 13 &= 1 \end{aligned}

195=3×5×13195 = 3 \times 5 \times 13

For 6565:

65÷5=1313÷13=1\begin{aligned} 65 \div 5 &= 13 \\ 13 \div 13 &= 1 \end{aligned}

65=5×1365 = 5 \times 13

Step 2 · Calculate LCM

Taking the highest power of each prime factor (313^1, 515^1, 717^1, 13113^1):

LCM(105,195,65)=31×51×71×131=3×5×7×13=15×91=1365\begin{aligned} \operatorname{LCM}(105, 195, 65) &= 3^1 \times 5^1 \times 7^1 \times 13^1 \\ &= 3 \times 5 \times 7 \times 13 \\ &= 15 \times 91 \\ &= 1365 \end{aligned}
Answer

(c) 1365

(d) 222, 370

Step 1 · Prime Factorisation of 222 and 370

For 222222:

222÷2=111111÷3=3737÷37=1\begin{aligned} 222 \div 2 &= 111 \\ 111 \div 3 &= 37 \\ 37 \div 37 &= 1 \end{aligned}

222=2×3×37222 = 2 \times 3 \times 37

For 370370:

370÷2=185185÷5=3737÷37=1\begin{aligned} 370 \div 2 &= 185 \\ 185 \div 5 &= 37 \\ 37 \div 37 &= 1 \end{aligned}

370=2×5×37370 = 2 \times 5 \times 37

Step 2 · Calculate LCM

Taking the highest power of each prime factor (212^1, 313^1, 515^1, 37137^1):

LCM(222,370)=21×31×51×371=2×3×5×37=30×37=1110\begin{aligned} \operatorname{LCM}(222, 370) &= 2^1 \times 3^1 \times 5^1 \times 37^1 \\ &= 2 \times 3 \times 5 \times 37 \\ &= 30 \times 37 \\ &= 1110 \end{aligned}
Answer

(d) 1110

Common Mistakes
  • Lowest vs. Highest Power: Taking the lowest power of common factors calculates the HCF, not the LCM. Always take the highest power of every prime factor present.
  • Missing Non-Common Factors: For LCM, you must include factors that appear in only one number (e.g., 55 in part (a), 77 in part (c)), not just the common ones.

More questions in FIO

Q1

List all the factors of the following numbers:

(a) 90

(b) 105

(c) 132

(d) 360 (this number has 24 factors)

(e) 840 (this number has 32 factors)

Q2

Find the common factors and the HCF of the following numbers:

(a) 50, 60

(b) 140, 275

(c) 77, 725

(d) 370, 592

(e) 81, 243

Q3

How do we directly find the HCF without listing all the factors?

Q4

Find the HCF of the following numbers:

(a) 24, 180

(b) 42, 75, 24

(c) 240, 378

(d) 400, 2500

(e) 300, 800

Q5

Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72=6×1272 = 6 \times 12 and 144=8×18144 = 8 \times 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?

Q6

Find the LCM of the following numbers:

(a) 30, 72

(b) 36, 54

(c) 105, 195, 65

(d) 222, 370

Q7

Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold.

(a) Two consecutive even numbers

(b) Two consecutive odd numbers

(c) Two even numbers

(d) Two consecutive numbers

(e) Two co-prime numbers

Share your observations with the class.

Q8

The LCM of 3 and 24 is 24 (it is one of the two given numbers).

(a) Find more such number pairs where the LCM is one of the two numbers.

(b) Make a general statement about such numbers. Describe such number pairs using algebra.

Q9

Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold.

(a) Two multiples of 3

(b) Two consecutive even numbers

(c) Two consecutive numbers

(d) Two co-prime numbers

Q10

In the two rows below, colours repeat as shown. When will the blue stars meet next?

Q11

(a) Is 5×7×11×115 \times 7 \times 11 \times 11 a multiple of 5×7×7×11×25 \times 7 \times 7 \times 11 \times 2?

(b) Is 5×7×11×115 \times 7 \times 11 \times 11 a factor of 5×7×7×11×25 \times 7 \times 7 \times 11 \times 2?

Q12

Find the HCF and LCM of the following (state your answers in the form of prime factorisations):

(a) 3×3×5×7×73 \times 3 \times 5 \times 7 \times 7 and 12×7×1112 \times 7 \times 11

(b) 45 and 36

Q13

Find two numbers whose HCF is 1 and LCM is 66.

Q14

A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)

Q15

The length, width, and height of a box are 12 cm12\text{ cm}, 18 cm18\text{ cm}, and 36 cm36\text{ cm} respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm9\text{ cm}

(b) 6 cm6\text{ cm}

(c) 4 cm4\text{ cm}

(d) 3 cm3\text{ cm}

(e) 2 cm2\text{ cm}

Q16

Among the numbers below, which is the largest number that perfectly divides both 306 and 36?

(a) 36

(b) 612

(c) 18

(d) 3

(e) 2

(f) 360

Q17

Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Q18

Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?

(a) 72

(b) 90

(c) 45

(d) 3

(e) 36

(f) None of these

Q19

Tick the correct statement(s). The LCM of two different prime numbers (m,nm, n) can be:

(a) Less than both numbers

(b) In between the two numbers

(c) Greater than both numbers

(d) Less than m×nm \times n

(e) Greater than m×nm \times n

Q20

A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?

Q21

What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?

Q22

Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 815\dfrac{8}{15}, 120\dfrac{1}{20}, 736\dfrac{7}{36}, 1163\dfrac{11}{63} and 121\dfrac{1}{21}. What do you get? How can we find this sum efficiently?

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