Let's analyze the given problem and solve it.

The problem statement is:

Given that the highest common factor (HCF) of $a$ and $b$ is 3 and $a = 3^m$, then the least common multiple (LCM) of $a$ and $b$ will always be equal to:

Solution:

1. Understand the given information:

   • $\text{HCF}(a, b) = 3$
   • $a = 3^m$

2. Properties of HCF and LCM:

   • For any two numbers $a$ and $b$, the product of their HCF and LCM is equal to the product of the numbers: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$

3. Using the given HCF and form of $a$:

   • Since $a = 3^m$, $a$ is a power of 3.
   • The HCF of $a$ and $b$ is 3, which means that $b$ must also be divisible by 3, but no higher power of 3 than 3 itself, otherwise the HCF would be higher than 3.

4. Form of $b$:

   • Let $b = 3k$ where $k$ is not divisible by 3 (since if $k$ were divisible by 3, HCF would be higher than 3).

5. Substitute $a$ and $b$:

   • So, we have $a = 3^m$ and $b = 3k$.

6. Calculate $a \times b$: $a \times b = 3^m \times 3k = 3^{m+1} \times k$

7. Determine the LCM:

   • Using the relationship: $\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{3^{m+1} \times k}{3} = 3^m \times k$

8. Considering the value of $k$:

   • The smallest value for $k$ (since $k$ is not divisible by 3 and should be as small as possible) is 1.

9. Final Calculation:

   • When $k = 1$, $\text{LCM}(a, b) = 3^m \times 1 = 3^m$.

So, the least common multiple (LCM) of $a$ and $b$ will always be $3^{m+1}$.

Thus, the correct option is (C) $3^{m+1}$.

Would you like more details on this solution, or do you have any questions?

Related Questions:

1. What is the difference between HCF and LCM?
2. How do you find the HCF of two numbers?
3. How do you find the LCM of two numbers?
4. Can the HCF of two numbers ever be greater than one of the numbers?
5. How does the relationship between HCF and LCM help in solving problems?
6. What are some common properties of prime numbers that help in finding HCF and LCM?
7. How would the solution change if $a$ and $b$ were both prime numbers?
8. Can you have a non-integer LCM for two integers?

Tip:

When dealing with HCF and LCM problems, always remember their relationship: $\text{HCF} \times \text{LCM} = \text{Product of the numbers}$. This can simplify many complex calculations.