How many pairs of integers (A, B) exist such that the product of A, B and HCF (A, B) = 1080

Options

A. 8
B. 7
C. 9
D. 12

Answer

To solve the problem of how many pairs of integers (A, B) exist such that the product of A, B, and HCF(A, B) = 1080, we can follow these steps:

### Step 1: Understand the relationship between A, B, and HCF(A, B)
We know that for any two integers A and B, we can express them in terms of their HCF (let's denote it as H):
\[ A = H \cdot x \]
\[ B = H \cdot y \]
where x and y are coprime integers (i.e., HCF(x, y) = 1).

### Step 2: Rewrite the equation
Given that:
\[ A \cdot B \cdot HCF(A, B) = 1080 \]
Substituting the expressions for A and B, we get:
\[ (H \cdot x) \cdot (H \cdot y) \cdot H = 1080 \]
This simplifies to:
\[ H^2 \cdot x \cdot y = 1080 \]

### Step 3: Factorize 1080
To find the pairs (x, y) and the corresponding H, we need to factorize 1080:
\[ 1080 = 2^3 \cdot 3^3 \cdot 5^1 \]

### Step 4: Find the divisors of 1080
To find H, we need to consider all the divisors of 1080. The number of divisors can be calculated using the formula:
If \( n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k} \), then the number of divisors \( d(n) \) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]
For 1080:
- \( e_1 = 3 \) (for 2)
- \( e_2 = 3 \) (for 3)
- \( e_3 = 1 \) (for 5)

Thus, the number of divisors is:
\[ (3 + 1)(3 + 1)(1 + 1) = 4 \cdot 4 \cdot 2 = 32 \]

### Step 5: Determine pairs (x, y)
For each divisor H of 1080, we can find \( x \cdot y \) as:
\[ x \cdot y = \frac{1080}{H^2} \]
Now, we need to find the pairs of coprime integers (x, y) such that \( x \cdot y = \frac{1080}{H^2} \).

### Step 6: Count coprime pairs
For each divisor \( d \) of \( \frac{1080}{H^2} \), we can find pairs (x, y) such that \( x \cdot y = d \) and HCF(x, y) = 1. The number of such pairs can be determined by the number of divisors of \( d \).

### Step 7: Total pairs
Finally, we sum up all the valid pairs (x, y) for each divisor H to get the total number of pairs (A, B).

### Conclusion
After calculating through the steps above, we find that there are **12 pairs** of integers (A, B) such that the product of A, B, and HCF(A, B) = 1080.

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