IF HCF of two numbers is 4 and sum of both number is 36. How mony pair of numbers are possible .

Answer

To solve the problem, we need to find pairs of numbers \( (a, b) \) such that:

1. The Highest Common Factor (HCF) of \( a \) and \( b \) is 4.
2. The sum of \( a \) and \( b \) is 36.

### Step 1: Express the numbers in terms of their HCF
Since the HCF of \( a \) and \( b \) is 4, we can express \( a \) and \( b \) as:
\[ a = 4m \quad \text{and} \quad b = 4n \]
where \( m \) and \( n \) are coprime integers (i.e., their HCF is 1).

### Step 2: Set up the equation for the sum
Given that the sum of \( a \) and \( b \) is 36, we can write:
\[ a + b = 36 \]
Substituting the expressions for \( a \) and \( b \):
\[ 4m + 4n = 36 \]
Dividing the entire equation by 4:
\[ m + n = 9 \]

### Step 3: Find pairs of coprime integers
Now we need to find pairs of integers \( (m, n) \) such that:
1. \( m + n = 9 \)
2. \( \text{HCF}(m, n) = 1 \)

We can list the pairs of integers that sum to 9:
- \( (1, 8) \)
- \( (2, 7) \)
- \( (3, 6) \)
- \( (4, 5) \)
- \( (5, 4) \)
- \( (6, 3) \)
- \( (7, 2) \)
- \( (8, 1) \)

### Step 4: Check for coprimeness
Now we check which of these pairs are coprime:
- \( (1, 8) \) → HCF = 1 (Coprime)
- \( (2, 7) \) → HCF = 1 (Coprime)
- \( (3, 6) \) → HCF = 3 (Not coprime)
- \( (4, 5) \) → HCF = 1 (Coprime)
- \( (5, 4) \) → HCF = 1 (Coprime)
- \( (6, 3) \) → HCF = 3 (Not coprime)
- \( (7, 2) \) → HCF = 1 (Coprime)
- \( (8, 1) \) → HCF = 1 (Coprime)

### Step 5: List the valid pairs
The valid coprime pairs \( (m, n) \) are:
1. \( (1, 8) \)
2. \( (2, 7) \)
3. \( (4, 5) \)
4. \( (7, 2) \)
5. \( (8, 1) \)

### Step 6: Count the pairs
Thus, the pairs of numbers \( (a, b) \) that satisfy the conditions are:
1. \( (4 \times 1, 4 \times 8) = (4, 32) \)
2. \( (4 \times 2, 4 \times 7) = (8, 28) \)
3. \( (4 \times 4, 4 \times 5) = (16, 20) \)
4. \( (4 \times 7, 4 \times 2) = (28, 8) \)
5. \( (4 \times 8, 4 \times 1) = (32, 4) \)

### Conclusion
Thus, there are a total of **5 pairs of numbers** possible.

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