Find the number of pairs of two numbers whose HCF is 5 and their sum is 50.

Answer

To find the number of pairs of two numbers whose HCF is 5 and their sum is 50, we can follow these steps:

### Step 1: Understand the problem
We need to find two numbers, say \( a \) and \( b \), such that:
- The highest common factor (HCF) of \( a \) and \( b \) is 5.
- The sum of \( a \) and \( b \) is 50.

### Step 2: Express the numbers in terms of their HCF
Since the HCF is 5, we can express \( a \) and \( b \) as:
- \( a = 5m \)
- \( b = 5n \)
where \( m \) and \( n \) are coprime integers (i.e., their HCF is 1).

### Step 3: Set up the equation for the sum
From the problem, we know:
\[ a + b = 50 \]
Substituting the expressions for \( a \) and \( b \):
\[ 5m + 5n = 50 \]
Dividing the entire equation by 5 gives:
\[ m + n = 10 \]

### Step 4: Find pairs of coprime integers \( (m, n) \)
We need to find pairs of integers \( (m, n) \) such that:
1. \( m + n = 10 \)
2. \( \text{HCF}(m, n) = 1 \)

### Step 5: List the pairs that satisfy the conditions
The pairs \( (m, n) \) that add up to 10 are:
- \( (1, 9) \)
- \( (2, 8) \)
- \( (3, 7) \)
- \( (4, 6) \)
- \( (5, 5) \)

Now, we check which of these pairs are coprime:
- \( (1, 9) \): HCF is 1 (coprime)
- \( (2, 8) \): HCF is 2 (not coprime)
- \( (3, 7) \): HCF is 1 (coprime)
- \( (4, 6) \): HCF is 2 (not coprime)
- \( (5, 5) \): HCF is 5 (not coprime)

The valid coprime pairs are:
- \( (1, 9) \)
- \( (3, 7) \)

### Step 6: Count the pairs
Each valid pair \( (m, n) \) corresponds to a pair \( (a, b) \):
- For \( (1, 9) \): \( (5 \times 1, 5 \times 9) = (5, 45) \)
- For \( (3, 7) \): \( (5 \times 3, 5 \times 7) = (15, 35) \)

Thus, we have the pairs:
1. \( (5, 45) \)
2. \( (15, 35) \)

### Final Result
The total number of pairs of two numbers whose HCF is 5 and their sum is 50 is **2**.

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