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

We will find the Least Common Multiple (LCM) for each set of numbers.

Step 1 — Prime Factorization of 30 and 72

Let us find the prime factors for each number. We divide by prime numbers until we reach 1.

For 30:

30÷2=1530 \div 2 = 15

15÷3=515 \div 3 = 5

5÷5=15 \div 5 = 1

So, the prime factors of 30 are:

30=2×3×5\boxed{30 = 2 \times 3 \times 5}

For 72:

72÷2=3672 \div 2 = 36

36÷2=1836 \div 2 = 18

18÷2=918 \div 2 = 9

9÷3=39 \div 3 = 3

3÷3=13 \div 3 = 1

So, the prime factors of 72 are:

72=2×2×2×3×3=23×32\boxed{72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2}

Step 2 — Finding LCM(30, 72)

We take each prime factor with its highest power. The prime factors involved are 2, 3, and 5.

The highest power of 2 is 232^3 (from 72). The highest power of 3 is 323^2 (from 72). The highest power of 5 is 515^1 (from 30).

Now we multiply these highest powers together.

LCM(30,72)=23×32×51LCM(30, 72) = 2^3 \times 3^2 \times 5^1

=8×9×5= 8 \times 9 \times 5

=72×5= 72 \times 5

360\boxed{360}

Step 3 — Prime Factorization of 36 and 54

Let us find the prime factors for each number.

For 36:

36÷2=1836 \div 2 = 18

18÷2=918 \div 2 = 9

9÷3=39 \div 3 = 3

3÷3=13 \div 3 = 1

So, the prime factors of 36 are:

36=2×2×3×3=22×32\boxed{36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2}

For 54:

54÷2=2754 \div 2 = 27

27÷3=927 \div 3 = 9

9÷3=39 \div 3 = 3

3÷3=13 \div 3 = 1

So, the prime factors of 54 are:

54=2×3×3×3=21×33\boxed{54 = 2 \times 3 \times 3 \times 3 = 2^1 \times 3^3}

Step 4 — Finding LCM(36, 54)

We take each prime factor with its highest power. The prime factors involved are 2 and 3.

The highest power of 2 is 222^2 (from 36). The highest power of 3 is 333^3 (from 54).

Now we multiply these highest powers together.

LCM(36,54)=22×33LCM(36, 54) = 2^2 \times 3^3

=4×27= 4 \times 27

108\boxed{108}

Step 5 — Prime Factorization of 105, 195, and 65

Let us find the prime factors for each number.

For 105:

105÷3=35105 \div 3 = 35

35÷5=735 \div 5 = 7

7÷7=17 \div 7 = 1

So, the prime factors of 105 are:

105=3×5×7\boxed{105 = 3 \times 5 \times 7}

For 195:

195÷3=65195 \div 3 = 65

65÷5=1365 \div 5 = 13

13÷13=113 \div 13 = 1

So, the prime factors of 195 are:

195=3×5×13\boxed{195 = 3 \times 5 \times 13}

For 65:

65÷5=1365 \div 5 = 13

13÷13=113 \div 13 = 1

So, the prime factors of 65 are:

65=5×13\boxed{65 = 5 \times 13}

Step 6 — Finding LCM(105, 195, 65)

We take each prime factor with its highest power. The prime factors involved are 3, 5, 7, and 13.

The highest power of 3 is 313^1 (from 105 and 195). The highest power of 5 is 515^1 (from 105, 195, and 65). The highest power of 7 is 717^1 (from 105). The highest power of 13 is 13113^1 (from 195 and 65).

Now we multiply these highest powers together.

LCM(105,195,65)=31×51×71×131LCM(105, 195, 65) = 3^1 \times 5^1 \times 7^1 \times 13^1

=3×5×7×13= 3 \times 5 \times 7 \times 13

=15×91= 15 \times 91

1365\boxed{1365}

Step 7 — Prime Factorization of 222 and 370

Let us find the prime factors for each number.

For 222:

222÷2=111222 \div 2 = 111

111÷3=37111 \div 3 = 37

37÷37=137 \div 37 = 1

So, the prime factors of 222 are:

222=2×3×37\boxed{222 = 2 \times 3 \times 37}

For 370:

370÷2=185370 \div 2 = 185

185÷5=37185 \div 5 = 37

37÷37=137 \div 37 = 1

So, the prime factors of 370 are:

370=2×5×37\boxed{370 = 2 \times 5 \times 37}

Step 8 — Finding LCM(222, 370)

We take each prime factor with its highest power. The prime factors involved are 2, 3, 5, and 37.

The highest power of 2 is 212^1 (from 222 and 370). The highest power of 3 is 313^1 (from 222). The highest power of 5 is 515^1 (from 370). The highest power of 37 is 37137^1 (from 222 and 370).

Now we multiply these highest powers together.

LCM(222,370)=21×31×51×371LCM(222, 370) = 2^1 \times 3^1 \times 5^1 \times 37^1

=2×3×5×37= 2 \times 3 \times 5 \times 37

=30×37= 30 \times 37

1110\boxed{1110}

Answer

(a) The LCM of 30 and 72 is 360. (b) The LCM of 36 and 54 is 108. (c) The LCM of 105, 195, and 65 is 1365. (d) The LCM of 222 and 370 is 1110.

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 × 12 and 144 = 8 × 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 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm

(b) 6 cm

(c) 4 cm

(d) 3 cm

(e) 2 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\frac{8}{15}, 120\frac{1}{20}, 736\frac{7}{36}, 1163\frac{11}{63} and 121\frac{1}{21}. What do you get? How can we find this sum efficiently?

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