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

Understand the Question

In the game Fire in the Mountain, when a group number is called, players form groups of that size. If no one gets out, the total number of children ( $N$ ) is exactly divisible by that number.
If some people get out, $N$ is not divisible by that number.
Therefore, $N$ must be a common multiple of $6$ and $9$ (i.e., a multiple of $\text{LCM}(6, 9) = 18$ ), but not a multiple of $10$ .

Step 1 · Find the Common Multiple of 6 and 9

Let the initial number of children be $N$ .

No one got out when $6$ was called $\implies N$ is divisible by $6$ .
No one got out when $9$ was called $\implies N$ is divisible by $9$ .

Therefore, $N$ must be a multiple of $\text{LCM}(6, 9)$ . Diagram 1

Prime factorisation:

\begin{aligned} 6 &= 2 \times 3 \\ 9 &= 3 \times 3 = 3^2 \end{aligned}

\begin{aligned} \text{LCM}(6, 9) &= 2^1 \times 3^2 \\ &= 2 \times 9 \\ &= 18 \end{aligned}

Thus, $N$ must be a multiple of $18$ ( $18, 36, 54, 72, 90, \dots$ ).

Step 2 · Check the Options Against the Non-Divisibility Rule

Since some people got out when $10$ was called, $N$ is not divisible by $10$ .

Checking each option:

(a) 72: $72 \div 18 = 4 \quad (\text{Multiple of } 18)$ $72 \div 10 = 7 \text{ with remainder } 2 \quad (\text{Not divisible by } 10)$ $\rightarrow$ Possible
(b) 90: $90 \div 18 = 5 \quad (\text{Multiple of } 18)$ $90 \div 10 = 9 \quad (\text{Divisible by } 10)$ $\rightarrow$ Not possible
(c) 45: $45 \div 18 = 2 \text{ with remainder } 9 \quad (\text{Not a multiple of } 18)$ $\rightarrow$ Not possible
(d) 3: $3 < 18 \quad (\text{Not a multiple of } 18)$ $\rightarrow$ Not possible
(e) 36: $36 \div 18 = 2 \quad (\text{Multiple of } 18)$ $36 \div 10 = 3 \text{ with remainder } 6 \quad (\text{Not divisible by } 10)$ $\rightarrow$ Possible

Answer

(a) 72 and (e) 36

Common Mistakes

Including 90: $90$ is a multiple of $18$ , but since $90$ is completely divisible by $10$ , no one would get out when $10$ is called.
Confusing Factors and Multiples: Selecting $3$ because it is a common factor of $6$ and $9$ , rather than finding a common multiple ( $N \ge 18$ ).

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: $72 = 6 \times 12$ and $144 = 8 \times 18$ . 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 $12\text{ cm}$ , $18\text{ cm}$ , and $36\text{ cm}$ respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) $9\text{ cm}$

(b) $6\text{ cm}$

(d) $3\text{ cm}$

(e) $2\text{ cm}$

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