Let $$f : N \rightarrow N$$ be a function such that $$f(m+n) = f(m) + f(n)$$ for every $$m, n \in N$$. If $$f(6) = 18$$ then $$f(2) \cdot f(3)$$ is equal to:

We are given a function $$f : \mathbb N \rightarrow \mathbb N$$ satisfying the additive property

$$f(m+n)=f(m)+f(n) \quad \text{for all } m,n\in\mathbb N.$$

First, we recall the standard fact that if a function on the natural numbers is additive in this way, then its value at any positive integer can be built up by repeatedly adding the value at $$1$$. More precisely, we use the principle of mathematical induction:

For $$n=1$$ the statement $$f(1)=f(1)$$ is trivially true. Now assume $$f(k)=k\,f(1)$$ for some $$k\in\mathbb N$$. Then, using the given property,

$$f(k+1)=f(k)+f(1).$$

Substituting the induction hypothesis $$f(k)=k\,f(1)$$, we obtain

$$f(k+1)=k\,f(1)+f(1)=(k+1)\,f(1).$$

Thus by induction the general formula

$$f(n)=n\,f(1) \quad\text{holds for every } n\in\mathbb N.$$

Now we use the given numerical information. We are told that

$$f(6)=18.$$

Applying the formula $$f(6)=6\,f(1)$$ we write

$$6\,f(1)=18.$$

Dividing both sides by $$6$$, we get

$$f(1)=\frac{18}{6}=3.$$

With $$f(1)$$ known, we can find the required values:

For $$n=2$$,

$$f(2)=2\,f(1)=2\times 3=6.$$

For $$n=3$$,

$$f(3)=3\,f(1)=3\times 3=9.$$

The question asks for the product $$f(2)\cdot f(3)$$. Substituting the values just found,

$$f(2)\cdot f(3)=6\times 9=54.$$

Hence, the correct answer is Option A.