If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct?

The correct option is C - Both I and II.

Solution

Let the two distinct numbers be x and y.
According to the fundamental property of numbers:
$HCF(x,y)\times LCM(x,y)=x\times y$

The question states that the product of HCF and LCM is the cube of one of the numbers. Let that number be x.
So, $HCF\times LCM=x^3$

Equating the two expressions:
$x\times y=x^3$
Dividing both sides by x (since x is a number and cannot be zero):
$y=x^2$

This tells us that one number (y) is the square of the other number (x). Since the numbers are distinct, x cannot be 1 (because if x=1, y=1^2=1, making them identical).

Testing Statement II: One of the numbers is a perfect square.
Since $y=x^2$, y is definitely a perfect square. Thus, Statement II is correct.

Testing Statement I: The difference of the numbers is an even number.
The difference is $(y-x)$ or $(x^2-x)$.
$x^2-x$ can be written as $x(x-1)$.
In mathematics, the product of any two consecutive integers (x and x-1) is always even, because one of them must be even.
Example 1: If x=2, y=4. Difference = $4-2=2$ (Even).
Example 2: If x=3, y=9. Difference = $9-3=6$ (Even).
Example 3: If x=4, y=16. Difference = $16-4=12$ (Even).
Thus, Statement I is correct.

Why the other options are incorrect

• Option (a) - I only: This is incorrect because it ignores the fact that Statement II is also mathematically proven to be true ($y=x^2$).
• Option (b) - II only: This is incorrect because it overlooks the algebraic property that the difference between a square and its root $(x^2-x)$ is always an even number.
• Option (d) - Neither I nor II: This is incorrect because both statements are logically and mathematically derived from the given condition.

Key Concept

The relationship between HCF, LCM, and the product of two numbers ($HCF\times LCM=Product$), combined with the property that the product of consecutive integers is always even.