To solve the problem of finding the number of pairs of two numbers whose product is 300 and their HCF is 5, we can follow these steps:
### Step 1: Understand the relationship between the numbers
Let the two numbers be \( x \) and \( y \). We know that:
- The product of the two numbers is given by:
\[
x \times y = 300
\]
- The highest common factor (HCF) of the two numbers is given as:
\[
\text{HCF}(x, y) = 5
\]
### Step 2: Express the numbers in terms of their HCF
Since the HCF of \( x \) and \( y \) is 5, we can express \( x \) and \( y \) as:
\[
x = 5a \quad \text{and} \quad y = 5b
\]
where \( a \) and \( b \) are coprime (i.e., their HCF is 1).
### Step 3: Substitute into the product equation
Substituting \( x \) and \( y \) into the product equation gives:
\[
(5a) \times (5b) = 300
\]
This simplifies to:
\[
25ab = 300
\]
### Step 4: Solve for \( ab \)
Dividing both sides by 25:
\[
ab = \frac{300}{25} = 12
\]
### Step 5: Find the pairs of \( (a, b) \)
Next, we need to find pairs of integers \( (a, b) \) such that \( ab = 12 \). The pairs of factors of 12 are:
- \( (1, 12) \)
- \( (2, 6) \)
- \( (3, 4) \)
### Step 6: Count the pairs
Since \( a \) and \( b \) can be swapped, we need to consider both arrangements:
1. \( (1, 12) \) and \( (12, 1) \)
2. \( (2, 6) \) and \( (6, 2) \)
3. \( (3, 4) \) and \( (4, 3) \)
This gives us a total of 6 pairs:
- \( (1, 12) \)
- \( (12, 1) \)
- \( (2, 6) \)
- \( (6, 2) \)
- \( (3, 4) \)
- \( (4, 3) \)
### Final Answer
Thus, the number of pairs of two numbers whose product is 300 and their HCF is 5 is **6**.
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