Finding Common Ground | IT

Question 26

Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product. Do you see any pattern? Use these numbers:

(a) 45, 105

(b) 275, 352

Solution

Understand the Question

For any two positive integers $A$ and $B$ , their product is related to their $\text{LCM}$ and $\text{HCF}$ by the fundamental identity: $\text{Product of numbers} = A \times B = \text{LCM}(A, B) \times \text{HCF}(A, B)$
Since $\text{Product} = \text{LCM} \times \text{HCF}$ , the $\text{LCM}$ divides the product completely with no remainder, meaning $\text{LCM}$ is always a factor of the product.
The number by which the $\text{LCM}$ must be multiplied to obtain the product is precisely the $\mathbf{\text{HCF}}$ of the two numbers: $\text{Multiplier} = \dfrac{\text{Product}}{\text{LCM}} = \text{HCF}$

(a) 45, 105

Step 1 · Find Prime Factorisation and Product

Prime factorisation of the numbers:

\begin{aligned} 45 &= 3 \times 3 \times 5 = 3^2 \times 5^1 \\ 105 &= 3 \times 5 \times 7 = 3^1 \times 5^1 \times 7^1 \end{aligned}

Product of the numbers:

\begin{aligned} \text{Product} &= 45 \times 105 \\ &= 4725 \end{aligned}

Step 2 · Calculate LCM, Multiplier, and HCF

Taking the highest powers of all prime factors to find $\text{LCM}$ :

\begin{aligned} \text{LCM}(45, 105) &= 3^2 \times 5^1 \times 7^1 \\ &= 9 \times 5 \times 7 \\ &= 315 \end{aligned}

Check if $\text{LCM}$ is a factor of the product by dividing:

\begin{aligned} \text{Multiplier} &= \dfrac{\text{Product}}{\text{LCM}} \\[0.6em] &= \dfrac{4725}{315} \\[0.6em] &= 15 \end{aligned}

Taking the lowest powers of common prime factors to find $\text{HCF}$ :

\begin{aligned} \text{HCF}(45, 105) &= 3^1 \times 5^1 \\ &= 15 \end{aligned}

Since the remainder is $0$ , $\text{LCM}$ is a factor of the product, and the multiplier equals $\text{HCF} = 15$ .

Answer

(a) Yes, $\text{LCM}$ is a factor of the product. The multiplier is $15$ .

(b) 275, 352

Step 1 · Find Prime Factorisation and Product

Prime factorisation of the numbers:

\begin{aligned} 275 &= 5 \times 5 \times 11 = 5^2 \times 11^1 \\ 352 &= 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^5 \times 11^1 \end{aligned}

Product of the numbers:

\begin{aligned} \text{Product} &= 275 \times 352 \\ &= 96800 \end{aligned}

Step 2 · Calculate LCM, Multiplier, and HCF

Taking the highest powers of all prime factors to find $\text{LCM}$ :

\begin{aligned} \text{LCM}(275, 352) &= 2^5 \times 5^2 \times 11^1 \\ &= 32 \times 25 \times 11 \\ &= 800 \times 11 \\ &= 8800 \end{aligned}

Check if $\text{LCM}$ is a factor of the product by dividing:

\begin{aligned} \text{Multiplier} &= \dfrac{\text{Product}}{\text{LCM}} \\[0.6em] &= \dfrac{96800}{8800} \\[0.6em] &= 11 \end{aligned}

Taking the lowest power of common prime factors to find $\text{HCF}$ :

\begin{aligned} \text{HCF}(275, 352) &= 11^1 \\ &= 11 \end{aligned}

Since the remainder is $0$ , $\text{LCM}$ is a factor of the product, and the multiplier equals $\text{HCF} = 11$ .

Answer

(b) Yes, $\text{LCM}$ is a factor of the product. The multiplier is $11$ .

(c) 222, 370

Step 1 · Find Prime Factorisation and Product

Prime factorisation of the numbers:

\begin{aligned} 222 &= 2 \times 3 \times 37 = 2^1 \times 3^1 \times 37^1 \\ 370 &= 2 \times 5 \times 37 = 2^1 \times 5^1 \times 37^1 \end{aligned}

Product of the numbers:

\begin{aligned} \text{Product} &= 222 \times 370 \\ &= 82140 \end{aligned}

Step 2 · Calculate LCM, Multiplier, and HCF

Taking the highest powers of all prime factors to find $\text{LCM}$ :

\begin{aligned} \text{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}

Check if $\text{LCM}$ is a factor of the product by dividing:

\begin{aligned} \text{Multiplier} &= \dfrac{\text{Product}}{\text{LCM}} \\[0.6em] &= \dfrac{82140}{1110} \\[0.6em] &= 74 \end{aligned}

Taking the lowest powers of common prime factors to find $\text{HCF}$ :

\begin{aligned} \text{HCF}(222, 370) &= 2^1 \times 37^1 \\ &= 2 \times 37 \\ &= 74 \end{aligned}

Since the remainder is $0$ , $\text{LCM}$ is a factor of the product, and the multiplier equals $\text{HCF} = 74$ .

Answer

(c) Yes, $\text{LCM}$ is a factor of the product. The multiplier is $74$ .

Common Mistakes

Confusing LCM and HCF Powers: Remember that $\text{LCM}$ takes the highest power of every prime factor present, while $\text{HCF}$ takes the lowest power of only the common prime factors.
Applying the Property to 3 Numbers: The property $\text{Product} = \text{LCM} \times \text{HCF}$ is valid only for two numbers, not for three or more numbers.

More questions in IT

Context: Sameeksha is building her new house. The main room of the house is $12\text{ ft}$ by $16\text{ ft}$ . She wants to cover the floor with square tiles of the same size, using as few tiles as possible, with the length of the tile being a whole number of feet. She needs tiles of size $4\text{ ft}$ .

Q. How many tiles of this size should she purchase?

What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change?

Context: Lekhana bought $84\text{ kg}$ of rice from one farm and $108\text{ kg}$ from another. She wants to pack them in bags of equal weight (whole number of $\text{kg}$ ) using as few bags as possible. The common factors of $84$ and $108$ are $1, 2, 3, 4, 6$ , and $12$ .

Q. Which weight should she choose to minimise the number of bags?

Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the chapter ‘Prime Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and tries to collect the treasure. In each case below, the two numbers upon which treasures are kept are given. Find the longest jump size (starting from 0) using which Jumpy can land on both the numbers having the treasure.

(a) 14 and 30

(b) 7 and 11

(d) 28 and 42

Is the longest jump size for the numbers the same as their HCF? Explain why it is so.

Can this process be simplified? Can it be made more reliable?

Can you see what is happening below?

Can you write the prime factorisation of $105$ and $30$ using these two figures?

Try finding the prime factorisation of 1200 using the method above. If we had used the earlier method, our calculation would have been as follows:

$1200 = 40 \times 30 = 5 \times 8 \times 5 \times 6 = \dots$

Which calculation is easier to carry out?

Context: Consider the number 840 and its prime factorisation $2 \times 2 \times 2 \times 3 \times 5 \times 7$ .

Q. Is $2 \times 2 \times 7 = 28$ a factor of 840?

Q10

Context: Consider the number 840 and its prime factorisation $2 \times 2 \times 2 \times 3 \times 5 \times 7$ .

Q. If yes, what should it be multiplied by to get 840?

Q11

Context: Consider the number 840 and its prime factorization $2 \times 2 \times 2 \times 3 \times 5 \times 7$ .

Similarly, is $2 \times 7 = 14$ a factor of 840? Why or why not?

Is $2 \times 2 \times 2$ a factor of 840? Why or why not?

Is $3 \times 3 \times 3$ a factor of 840? Why or why not?

Can we use this idea to list down all the possible factors of a number using just its prime factors?

Q12

Context: The factors of 225 are found to be 1, 3, 5, 9, 15, 25, 45, 75, 225.

Q. Check that all the factors of 225 occur in this list.

Q13

Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find the first number for which ‘idli-vada’ will be called out:

(a) 4 and 6

(b) 7 and 11

(d) 15 and 55

Is the answer always the LCM of the two numbers? Explain.

Q14

Context: Consider the numbers 14 and 35, with prime factorisations $14 = 2 \times 7$ and $35 = 5 \times 7$ .

Q. Is $2 \times 3 \times 5 \times 7$ also a common multiple?

Q15

Find more such number pairs where the HCF is one of the two numbers. How can we describe such pairs of numbers?

Q16

Context: If $n$ is a number, then any multiple of $n$ can be written as a positive integer multiplied by $n$ . For example, if we take $n$ and $5n$ (short for $5 \times n$ ), then $5n$ is a multiple of $n$ , and $n$ is a factor of $5n$ . The HCF of $n$ and $5n = n$ .

Q. For number pairs satisfying this property (i.e., one of the numbers is the HCF),

(a) if $m$ is a number, what could be the other number?

(b) if $7k$ is a number, what could be the other number?

Q17

What happens to the HCF of two numbers if both numbers are doubled? Take some pairs of numbers and explore. Are you able to see why the HCF will also double?

Q18

Here are some more numbers where both numbers are multiples of the same number. Find their HCF:

(a) $18 \times 10$ , $18 \times 15$

(b) $10 \times 38$ , $10 \times 21$

(d) $12 \times 16$ , $12 \times 20$

Q19

In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens.

Q20

Efficient Procedures for HCF and LCM

See the procedure on the right. Can you explain how it has been carried out?

Q21

How do we use this to find the HCF of 84 and 180? Explore.

[Hint: Observe that $84 = 2 \times 2 \times 3 \times 7$ , and $180 = 2 \times 2 \times 3 \times 15$ similar to prime factorisation]

Q22

Why are these the LCMs?

[Hint: Will the product of the factors marked as the LCM of 300 and 150 contain the prime factorisations of both 300 and 150? Is this the smallest such number?]

Q23

You can try this method for these pairs of numbers.

(a) 90 and 150

(b) 84 and 132

Q24

Property Involving both the HCF and the LCM

Which is greater — the LCM of two numbers or their product?

Q25

You could analyse the above statement using examples. Then try to reason or prove, why the LCM is never greater than the product of the numbers.

[Hint: Is the product also a common multiple of the two numbers?]

Q26

(a) 45, 105

(b) 275, 352

Q27

Context: Consider the pairs of numbers: (a) 45, 105; (b) 275, 352; (c) 222, 370.

Q. Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?

Q28

Why does this happen? Can you give an explanation or proof?

[Hint: Consider the prime factorisation of the given numbers. Among their prime factors, some are common to both factorisations, and the rest occur in only one of them. Between the HCF and the LCM, see how the common and non-common prime factors get distributed. In the product, observe how these two kinds of prime factors occur. Compare them.]

Q29

Explore whether this property holds when 3 numbers are considered.

Q30

Context: The largest prime found so far has 4,10,24,320 digits! It was discovered on October 12, 2024.

Q. If I start writing this number, how long could it take me?