Suppose that a function $$f : \mathbb{R} \to \mathbb{R}$$ satisfies $$f(x+y) = f(x)f(y)$$ for all $$x, y \in \mathbb{R}$$ and $$f(1) = 3$$. If $$\sum_{i=1}^{n} f(i) = 363$$, then $$n$$ is equal to_____.

We have a real-valued function $$f : \mathbb{R} \to \mathbb{R}$$ that obeys the functional equation $$f(x+y)=f(x)f(y)$$ for every pair of real numbers $$x$$ and $$y$$, and it is given that $$f(1)=3$$.

First we want to find the values of $$f(i)$$ for natural numbers $$i=1,2,3,\ldots$$. To do this, we repeatedly use the given rule $$f(x+y)=f(x)f(y)$$.

Take $$i=2$$. Write $$2=1+1$$, then

$$f(2)=f(1+1)=f(1)f(1).$$

Substituting the known value $$f(1)=3$$, we get

$$f(2)=3\cdot3=3^{2}.$$

Now take $$i=3$$. Write $$3=2+1$$, and apply the rule again:

$$f(3)=f(2+1)=f(2)f(1).$$

We already have $$f(2)=3^{2}$$ and $$f(1)=3$$, so

$$f(3)=3^{2}\cdot3=3^{3}.$$

Proceeding in the same manner, every time we increase the input by $$1$$ we multiply the previous value by $$3$$. Hence, by simple induction, we arrive at

$$f(i)=3^{i}\quad\text{for all natural numbers }i.$$

The problem states that

$$\sum_{i=1}^{n} f(i)=363.$$

Substituting $$f(i)=3^{i}$$ into the summation gives

$$\sum_{i=1}^{n} 3^{i}=363.$$

The left-hand side is a finite geometric series with first term $$3$$ and common ratio $$3$$. The standard formula for the sum of the first $$n$$ terms of a geometric progression with first term $$a$$, ratio $$r$$ $$(r\neq1)$$ is

$$S_n=\frac{a\,(r^{\,n}-1)}{r-1}.$$

Here $$a=3$$ and $$r=3$$, so

$$\sum_{i=1}^{n} 3^{i}= \frac{3\,\bigl(3^{\,n}-1\bigr)}{3-1} =\frac{3\,(3^{\,n}-1)}{2}.$$

We equate this to $$363$$ as demanded by the question:

$$\frac{3\,(3^{\,n}-1)}{2}=363.$$

Now we solve step by step. Multiply both sides by $$2$$:

$$3\,(3^{\,n}-1)=726.$$

Divide by $$3$$:

$$3^{\,n}-1=242.$$

Add $$1$$ to both sides:

$$3^{\,n}=243.$$

We recognize that $$243=3^{5}$$. Since the base $$3$$ is positive and not equal to $$1$$, equality of the powers implies equality of the exponents, giving

$$n=5.$$

So, the answer is $$5$$.