The product of two numbers is $1500$ and their HCF is $10$. How many such possible pairs of numbers are there?

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**Solution:**

1. Let the two numbers be $10x$ and $10y$, where $x$ and $y$ are co-prime (i.e., their HCF is 1).

2. Given the product of the two numbers is:
   $$(10x) 	imes (10y) = 1500$$
   $$100xy = 1500$$
   $$xy = \frac{1500}{100} = 15$$

3. Since $x$ and $y$ are co-prime and their product is 15, we need to find pairs of co-prime numbers whose product is 15.

4. The factor pairs of 15 are:
   - $(1, 15)$
   - $(3, 5)$
   - $(5, 3)$
   - $(15, 1)$

5. Among these, the pairs where $x$ and $y$ are co-prime are:
   - $(1, 15)$
   - $(3, 5)$

Note that $(5, 3)$ and $(15, 1)$ are just reversals of the above pairs.

6. Therefore, the number of distinct pairs $(10x, 10y)$ is 2.

**Answer:** 2

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