The sum of two numbers is 1215 and their HCF is 81. How many such pairs of numbers can be formed ?

Options

A. 3
B. 4
C. 6
D. None of these

Answer

To solve the problem, we need to find how many pairs of numbers can be formed given that their sum is 1215 and their HCF is 81.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We know that the sum of two numbers is 1215.
- Their Highest Common Factor (HCF) is 81.

2. **Expressing the Numbers:**
- Since the HCF of the two numbers is 81, we can express the two numbers as:
\[ a = 81x \quad \text{and} \quad b = 81y \]
- Here, \(x\) and \(y\) are co-prime numbers (they have no common factors other than 1).

3. **Setting Up the Equation:**
- According to the problem, the sum of the two numbers is:
\[ a + b = 1215 \]
- Substituting the expressions for \(a\) and \(b\):
\[ 81x + 81y = 1215 \]

4. **Simplifying the Equation:**
- We can factor out 81 from the left side:
\[ 81(x + y) = 1215 \]
- Now, divide both sides by 81:
\[ x + y = \frac{1215}{81} \]

5. **Calculating \(x + y\):**
- Performing the division:
\[ x + y = 15 \]

6. **Finding Pairs of \((x, y)\):**
- Now we need to find pairs of co-prime integers \(x\) and \(y\) such that:
\[ x + y = 15 \]
- The possible pairs \((x, y)\) that satisfy this equation are:
- (1, 14)
- (2, 13)
- (4, 11)
- (7, 8)

7. **Checking Co-primality:**
- We need to ensure that each pair is co-prime:
- (1, 14) → Co-prime
- (2, 13) → Co-prime
- (4, 11) → Co-prime
- (7, 8) → Co-prime

8. **Conclusion:**
- All four pairs are valid. Therefore, the total number of pairs of numbers that can be formed is **4**.

### Final Answer:
The number of such pairs of numbers that can be formed is **4**.

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