In CL-05, every time you computed HCF and LCM of two numbers, you verified by checking that:

HCF $\times$ LCM = Product of the two numbers

This was not a coincidence — it is a guaranteed mathematical property.

In this section, you will:

Understand WHY this property holds
Learn to use it systematically

And finally, we will understand the important caveat:

Number of Values	Does the Property Work?
Two Numbers	✅ Yes
Three or More	❌ No

1. Stating and Verifying the Product Property

The product property connects HCF and LCM in a beautiful way.

Before using it to solve problems, let's make sure you can:

State the property correctly.
Verify it using specific numbers.

📋 Given Info

Given Information

For the numbers 126 and 156:

HCF = $6$
LCM = $3276$

✍️ Question

Your Task 📝

State the product property for HCF and LCM.
Verify it using the given values for $126$ and $156$ .

The product property says:

HCF(a, b) × LCM(a, b) = a × b

✍️ Yes/No

Yes or No?

Does this property

HCF(a, b) \times LCM(a, b) = a \times b

apply to any two positive integers?

This holds for any two positive integers $a$ and $b$ . Let me verify with 126 and 156.

✍️ MCQ

Choose one

To verify the property for

126

and

156

, we need to check if

HCF \times LCM

equals:

Left side: HCF × LCM = 6 × 3276

Break it down:

$6 \times 3000 = 18000$
$6 \times 276 = 1656$

✍️ FIB

Fill in the blank

What is

18000 + 1656

?

$6 \times 3276 = 18000 + 1656 =$ 19656

Right side: a × b = 126 × 156

Break it down:

$126 \times 150 = 18900$
$126 \times 6 = 756$

✍️ MCQ

Choose one

What should

126 \times 156

equal to verify the property?

$126 \times 156 = 18900 + 756 =$ 19656

Both sides = 19656. Match! ✓

Why does this always work?

To understand why $H C F \times L C M = a \times b$ , we look at the prime factors of the numbers. For every prime factor $p$ :

HCF chooses the smallest power: $p^{\min (e_{a}, e_{b})}$
LCM chooses the largest power: $p^{\max (e_{a}, e_{b})}$

✍️ MCQ

Choose one

If

a = 2^3 \times \dots

and

b = 2^5 \times \dots

, what is the power of

2

in

HCF(a, b)

?

The Power of Addition

When we multiply HCF and LCM, we add their exponents:

HCF × LCM exponent $\to p^{(\min + \max)}$

When we multiply the original numbers $a$ and $b$ , we also add exponents:

$a \times b$ exponent $\to p^{e_{a}} \times p^{e_{b}} = p^{(e_{a} + e_{b})}$

The Identity: $\min (x, y) + \max (x, y) = x + y$

✍️ T/F

True or False?

Is the statement

\min(x, y) + \max(x, y) = x + y

always true?

Conclusion

Because the sum of the exponents matches for every prime factor:

HCF × LCM and $a \times b$ have the exact same prime factorisation.

Therefore, they must be equal!

Important Caveat

This property works for TWO numbers only.

✍️ Yes/No

Yes or No?

Can we use the formula

HCF(a, b, c) \times LCM(a, b, c) = a \times b \times c

for three numbers?

For three or more numbers, $\min (x, y, z) + \max (x, y, z)$ does NOT equal $x + y + z$ , so the property fails.

✍️ T/F

True or False?

The product of the HCF and LCM of any three positive integers is equal to the product of those three integers.

2. Why the Property Fails for Three Numbers

A Common Exam Trap ⚠️

A very common exam question tests whether you know the limitation of the product property.

The Mistake: Many students incorrectly extend this rule to three numbers.

Let's see if you can spot the issue!

✍️ Question

Check this claim 🔍

A student claims that HCF × LCM = a × b × c for any three numbers.

Given information:

Numbers ( $a, b, c$ ): $36$ , $60$ , and $84$
HCF: $12$
LCM: $1260$

Is the student correct?

(Compute both sides and compare)

Let me check the student's claim by computing both sides.

Left side: HCF × LCM = 12 × 1260 = 12 × 1000 + 12 × 260 = 12000 + 3120 = 15,120

✍️ FIB

Fill in the blank

If the property worked, what should

36 \times 60 \times 84

equal?

Right side: a × b × c = 36 × 60 × 84 = 36 × 60 = 2160 = 2160 × 84 = 2160 × 80 + 2160 × 4 = 172800 + 8640 = 181,440

15,120 is NOT equal to 181,440. Not even close!

So the student's claim is wrong. The product property HCF × LCM = a × b works for two numbers only.

Why does it fail?

For two numbers, the logic is simple: $\min (x, y) + \max (x, y) = x + y$ .

However, for three numbers, $\min (x, y, z) + \max (x, y, z)$ does NOT equal $x + y + z$ . Why? Because the middle value is lost.

✍️ MCQ

Choose one

If three numbers have exponents

2, 5,

and

8

, what is

\min(2, 5, 8) + \max(2, 5, 8)

?

Example: Take the exponents of prime $3$ in our three numbers:

$36$ has $3^{2}$ (exponent $2$ )
$60$ has $3^{1}$ (exponent $1$ )
$84$ has $3^{1}$ (exponent $1$ )

$\min (2, 1, 1) = 1, \max (2, 1, 1) = 2$ $So, \min + \max = 3$

But look at the actual sum: $2 + 1 + 1 = 4$

The middle values are lost!

✍️ FIB

Fill in the blank

In the set of exponents

(2, 1, 1)

, which value was lost?

The textbook explicitly warns:

'The above result is true for two numbers only.'

✍️ T/F

True or False?

The property

HCF \times LCM = \text{product of numbers}

holds true for a set of four numbers.

This is a common exam trap — the examiner tests whether you know this limitation.

✍️ Yes/No

Yes or No?

Can you use the product property to find the

LCM

of

10, 20,

and

30

?

Understanding and Verifying the Product Property

1. Stating and Verifying the Product Property

Given Information

Your Task 📝

Why does this always work?

The Power of Addition

Conclusion

Important Caveat

2. Why the Property Fails for Three Numbers

A Common Exam Trap ⚠️

Check this claim 🔍

Why does it fail?