Finding Common Ground | FIO

Question 18

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

(d) 3

(e) 36

(f) None of these

Solution

We need to find the number of children that fits all the given rules.

Step 1 — Find the common multiple

Let us call the initial number of children N. When 6 was called, no one got out. This means N is perfectly divisible by 6. When 9 was called, no one got out. This means N is perfectly divisible by 9. So, N must be a common multiple of 6 and 9. We need to find the Least Common Multiple (LCM) of 6 and 9. First, we find the prime factors of each number.

$6 = 2 \times 3$

$9 = 3 \times 3 = 3^2$

To find the LCM, we take the highest power of each prime factor.

$\text{LCM}(6, 9) = 2^1 \times 3^2$

$= 2 \times 9$

$\boxed{\text{LCM}(6, 9) = \mathbf{18}}$

This means the number of children N must be a multiple of 18.

Diagram 1

Step 2 — Check the non-divisibility rule

When 10 was called, some people got out. This means N is NOT perfectly divisible by 10. So, N must be a multiple of 18 but not a multiple of 10. Let us check the given options.

(a) 72 Is 72 a multiple of 18? $72 \div 18 = 4$ Yes, it is. Is 72 divisible by 10? $72 \div 10 = 7 \text{ with a remainder of } 2$ No, it is not. So, 72 is a possible number of children.

(b) 90 Is 90 a multiple of 18? $90 \div 18 = 5$ Yes, it is. Is 90 divisible by 10? $90 \div 10 = 9$ Yes, it is. This means no one would get out if 10 was called. So, 90 is not a possible number of children.

(c) 45 Is 45 a multiple of 18? $45 \div 18 = 2 \text{ with a remainder of } 9$ No, it is not. So, 45 is not a possible number of children.

(d) 3 Is 3 a multiple of 18? $3 \div 18 \text{ is not a whole number}$ No, it is not. So, 3 is not a possible number of children.

(e) 36 Is 36 a multiple of 18? $36 \div 18 = 2$ Yes, it is. Is 36 divisible by 10? $36 \div 10 = 3 \text{ with a remainder of } 6$ No, it is not. So, 36 is a possible number of children.

We found two options that fit all the rules.

Answer

(i) 36 (ii) 72

More questions in FIO

List all the factors of the following numbers:

(a) 90

(b) 105

(d) 360 (this number has 24 factors)

(e) 840 (this number has 32 factors)

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

(a) 50, 60

(b) 140, 275

(d) 370, 592

(e) 81, 243

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

Find the HCF of the following numbers:

(a) 24, 180

(b) 42, 75, 24

(d) 400, 2500

(e) 300, 800

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?

Find the LCM of the following numbers:

(a) 30, 72

(b) 36, 54

(d) 222, 370

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

(d) Two consecutive numbers

(e) Two co-prime numbers

Share your observations with the class.

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.

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

(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 $5 \times 7 \times 11 \times 11$ a multiple of $5 \times 7 \times 7 \times 11 \times 2$ ?

(b) Is $5 \times 7 \times 11 \times 11$ a factor of $5 \times 7 \times 7 \times 11 \times 2$ ?

Q12

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

(a) $3 \times 3 \times 5 \times 7 \times 7$ and $12 \times 7 \times 11$

(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

(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

(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

(a) 72

(b) 90

(d) 3

(e) 36

(f) None of these

Q19

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

(a) Less than both numbers

(b) In between the two numbers

(d) Less than $m \times n$

(e) Greater than $m \times n$

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 $\frac{8}{15}$ , $\frac{1}{20}$ , $\frac{7}{36}$ , $\frac{11}{63}$ and $\frac{1}{21}$ . What do you get? How can we find this sum efficiently?