To solve the problem, we need to find pairs of numbers whose product is 84 and whose highest common factor (HCF) is 2.
### Step-by-Step Solution:
1. **Understanding the relationship between the numbers:**
Let the two numbers be \( x \) and \( y \). According to the problem, we have:
\[
x \cdot y = 84
\]
and the HCF of \( x \) and \( y \) is 2.
2. **Expressing the numbers in terms of their HCF:**
Since the HCF is 2, we can express the numbers as:
\[
x = 2a \quad \text{and} \quad y = 2b
\]
where \( a \) and \( b \) are coprime (i.e., their HCF is 1).
3. **Substituting into the product equation:**
Substituting \( x \) and \( y \) into the product equation gives:
\[
(2a) \cdot (2b) = 84
\]
Simplifying this, we find:
\[
4ab = 84
\]
Dividing both sides by 4, we get:
\[
ab = 21
\]
4. **Finding pairs of coprime factors of 21:**
Now we need to find pairs of integers \( (a, b) \) such that \( ab = 21 \) and \( \text{HCF}(a, b) = 1 \). The pairs of factors of 21 are:
- \( (1, 21) \)
- \( (3, 7) \)
5. **Checking coprimality of the pairs:**
- For the pair \( (1, 21) \), the HCF is 1.
- For the pair \( (3, 7) \), the HCF is also 1.
6. **Calculating the corresponding pairs of \( (x, y) \):**
Now we can find the corresponding pairs of \( (x, y) \):
- From \( (1, 21) \):
\[
x = 2 \cdot 1 = 2, \quad y = 2 \cdot 21 = 42 \quad \Rightarrow \quad (2, 42)
\]
- From \( (3, 7) \):
\[
x = 2 \cdot 3 = 6, \quad y = 2 \cdot 7 = 14 \quad \Rightarrow \quad (6, 14)
\]
7. **Listing the pairs:**
The pairs of numbers that satisfy the conditions are:
- \( (2, 42) \)
- \( (6, 14) \)
### Conclusion:
Thus, there are **two pairs of numbers** whose product is 84 and whose HCF is 2: \( (2, 42) \) and \( (6, 14) \).
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