Two positive numbers have their HCF as 12 and their product as 6336. Choose the correct option for the number of pairs possible for the given numbers.A. 2B. 3C. 4D. 5

We are given two positive numbers with the following information:

1. Their Highest Common Factor (HCF) is 12.
2. Their product is 6336.

Let the two numbers be denoted by $a$ and $b$. The relationship between the HCF, LCM, and the product of two numbers is:

$$\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$$

From the problem, we know:

$$\text{HCF}(a,b)=12\text{and}a\times b=6336$$

So, we can write:

$$12\times \text{LCM}(a,b)=6336$$

Solving for the LCM:

$$\text{LCM}(a,b)=\frac{6336}{12}=528$$

Next, since $a$ and $b$ are divisible by 12 (because their HCF is 12), we can express $a$ and $b$ as:

$$a=12x\text{and}b=12y$$

where $x$ and $y$ are coprime numbers (i.e., their HCF is 1). The product $a\times b$ is:

$$12x\times 12y=6336$$

Simplifying this:

$$144xy=6336$$

Dividing both sides by 144:

$$xy=\frac{6336}{144}=44$$

So, we need to find the pairs of coprime numbers $x$ and $y$ such that their product is 44.

The factor pairs of 44 are:

$$(1,44),(2,22),(4,11)$$

For each of these pairs, we check if $x$ and $y$ are coprime:

- For $(1,44)$, $\text{HCF}(1,44)=1$, so they are coprime.
- For $(2,22)$, $\text{HCF}(2,22)=2$, so they are not coprime.
- For $(4,11)$, $\text{HCF}(4,11)=1$, so they are coprime.

Thus, the valid coprime pairs are $(1,44)$ and $(4,11)$. These pairs correspond to the values of $x$ and $y$, and since the numbers $a$ and $b$ are symmetric, we can switch $x$ and $y$ in each pair.

So, the possible pairs for $(a,b)$ are:

1. $a=12\times 1=12,b=12\times 44=528$
2. $a=12\times 44=528,b=12\times 1=12$
3. $a=12\times 4=48,b=12\times 11=132$
4. $a=12\times 11=132,b=12\times 4=48$

Thus, the number of pairs possible is 4.

The correct option is **C. 4**.