Two positive integers a and b have their HCF as ` h( h ne 1)` . How many such integers exist if `(a + b+ h) = 15` ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find two positive integers \( a \) and \( b \) such that their highest common factor (HCF) is \( h \) (where \( h \neq 1 \)), and the equation \( a + b + h = 15 \) holds true.

### Step-by-Step Solution:

1. **Understanding the Equation**:
We start with the equation:
\[ a + b + h = 15 \]
Rearranging gives us:
\[ a + b = 15 - h \]

2. **Constraints on \( h \)**:
Since \( h \) is the HCF of \( a \) and \( b \) and must be greater than 1, the possible values for \( h \) can be 2, 3, 4, ..., up to 14 (since \( h \) must be less than 15).

3. **Finding Possible Values for \( h \)**:
We will check each possible value of \( h \) to see if we can find corresponding values of \( a \) and \( b \) such that both are positive integers and their HCF is \( h \).

4. **Testing Values of \( h \)**:
- **For \( h = 2 \)**:
\[ a + b = 15 - 2 = 13 \]
Possible pairs: (2, 11), (4, 9), (6, 7), (8, 5), (10, 3), (12, 1).
HCF of all pairs is not 2, so no valid pairs.

- **For \( h = 3 \)**:
\[ a + b = 15 - 3 = 12 \]
Possible pairs: (3, 9), (6, 6), (9, 3).
HCF of (3, 9) is 3, and HCF of (6, 6) is 6, so valid pair is (3, 9).

- **For \( h = 4 \)**:
\[ a + b = 15 - 4 = 11 \]
Possible pairs: (4, 7), (8, 3), (10, 1).
HCF of all pairs is not 4, so no valid pairs.

- **For \( h = 5 \)**:
\[ a + b = 15 - 5 = 10 \]
Possible pairs: (5, 5).
HCF of (5, 5) is 5, so valid pair is (5, 5).

- **For \( h = 6 \)**:
\[ a + b = 15 - 6 = 9 \]
Possible pairs: (6, 3), (4, 5).
HCF of (6, 3) is 3, and HCF of (4, 5) is 1, so no valid pairs.

- **For \( h = 7 \)**:
\[ a + b = 15 - 7 = 8 \]
Possible pairs: (7, 1).
HCF of (7, 1) is 1, so no valid pairs.

- **For \( h = 8 \)**:
\[ a + b = 15 - 8 = 7 \]
Possible pairs: (8, -1), (6, 1), (5, 2).
HCF of all pairs is not 8, so no valid pairs.

- **For \( h = 9 \)**:
\[ a + b = 15 - 9 = 6 \]
Possible pairs: (9, -3), (6, 0), (3, 3).
HCF of (3, 3) is 3, so no valid pairs.

- **For \( h = 10 \)**:
\[ a + b = 15 - 10 = 5 \]
Possible pairs: (10, -5), (5, 0).
HCF of all pairs is not 10, so no valid pairs.

- **For \( h = 11 \)**:
\[ a + b = 15 - 11 = 4 \]
Possible pairs: (11, -7), (4, 0).
HCF of all pairs is not 11, so no valid pairs.

- **For \( h = 12 \)**:
\[ a + b = 15 - 12 = 3 \]
Possible pairs: (12, -9), (3, 0).
HCF of all pairs is not 12, so no valid pairs.

- **For \( h = 13 \)**:
\[ a + b = 15 - 13 = 2 \]
Possible pairs: (13, -11), (2, 0).
HCF of all pairs is not 13, so no valid pairs.

- **For \( h = 14 \)**:
\[ a + b = 15 - 14 = 1 \]
Possible pairs: (14, -13), (1, 0).
HCF of all pairs is not 14, so no valid pairs.

5. **Valid Combinations**:
The valid combinations we found are:
- \( (3, 9) \) with \( h = 3 \)
- \( (5, 5) \) with \( h = 5 \)

### Conclusion:
Thus, there are **2 valid combinations** of integers \( a \) and \( b \) such that their HCF \( h \) is greater than 1 and satisfies the equation \( a + b + h = 15 \).

### Final Answer:
**2**