We are given that the LCM (Least Common Multiple) and the HCF (Highest Common Factor) of two numbers are $620$ and $4$ respectively.

Step-by-step solution:

1. Relationship between LCM and HCF:
   For any two numbers $a$ and $b$, the product of the LCM and HCF of the numbers is equal to the product of the numbers themselves: $\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b$ Using the values from the problem: $620 \times 4 = a \times b$ $a \times b = 2480$

2. Expressing $a$ and $b$ in terms of the HCF:
   Let $a = 4x$ and $b = 4y$, where $x$ and $y$ are co-prime integers (i.e., their HCF is 1). This is because the HCF of $a$ and $b$ is 4. Substituting into the product equation: $(4x) \times (4y) = 2480$ $16xy = 2480$ $xy = \frac{2480}{16} = 155$

3. Finding pairs of $x$ and $y$:
   Now, $x$ and $y$ are co-prime factors of 155. The prime factorization of 155 is: $155 = 5 \times 31$ The co-prime pairs $(x, y)$ that multiply to 155 are: $(1, 155), (5, 31)$

4. Counting the pairs:
   For each co-prime pair $(x, y)$, we can assign $a = 4x$ and $b = 4y$ or $a = 4y$ and $b = 4x$. Hence, each pair can be arranged in two ways. Since there are 2 pairs of $x$ and $y$, we have: $2 \times 2 = 4$ Thus, there are 4 such pairs of numbers $a$ and $b$.

Answer:

There are 4 pairs of numbers that satisfy the given conditions.

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Would you like a detailed explanation on any part of this solution? Here are some related questions you might find useful:

1. How do you find the LCM and HCF of two numbers manually?
2. What is the prime factorization method, and how is it useful in LCM/HCF problems?
3. Why do we use the relationship $\text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b$?
4. How can you find co-prime pairs of a number?
5. How would the process change if the LCM and HCF values were different?

Tip: When dealing with LCM and HCF problems, always break the numbers into their prime factors—it simplifies the process and reveals hidden relationships!