Finding the remainder of a 6 degree polynomial and polynomial itself. Using some algebraic or graphical techniques. In a six-degree polynomial, say $f(x)$ we have, $$f(1)=1, f(2)=1/2, f(3)=1/3, f(4)=1/4, f(5)=1/5, f(6)=1/6, f(7)=1/7$$ Find $f(8)$ and the polynomial $f(x)$.
This problem comes all the way from the book of an Indian Author. I tried solving the problem using algebra (definitely not by substituting values in the polynomial and solving for SIX variables). According to me, this question is definitely having a more rational approach. I tried using algebraic techniques by manipulating $f(x)$ writing it in terms of a new polynomial say $h(x)$ but in vain. So can anyone of the bright minds out there could help me in attempting the question. It would be a great help.
 A: Consider the polynomial $g(x) = xf(x) - 1$. You have that $g$ is a polynomial of degree 7.
You also know that $$g(n) = nf(n) - 1 = n\cdot1/n - 1= 0,$$ 
for $n = 1, \ldots, 7$. Thus, $g(x)$ factorises as
$$g(x)= A(x - 1)\cdots(x - 7), \quad (*)$$
where $A$ is some real constant.
(We got the above since $g(x)$ is a degree 7 polynomial.)
Now, all we need to do is determine $A$. This is easy because the definition of $g(x)$ tells us that $g(0) = 0f(0) - 1 = -1$. Substituting this in $(*)$ gives us:
$$-1 = A(-1)\cdots(-7) = -5040A$$
$$\implies A = \dfrac{1}{5040}.$$

Now, we can retrieve $f(x)$ as $f(x) = \dfrac{1}{x}(g(x) + 1)$ or
$$f(x) = \dfrac{1}{x}\left(\dfrac{1}{5040}(x - 1)\cdots(x - 7) + 1\right).$$
(Note that the constant of the polynomial within the parenthesis is $0$ and thus, $f(x)$ will indeed turn out to be a polynomial.)

Calculating $f(8)$ is considerably easier as we get
$$\begin{align}f(8) &= \dfrac{1}{8}\left(\dfrac{1}{5040}(8 - 1)\cdots(8 - 7) + 1\right)\\
&= \dfrac{1}{8}\left(\dfrac{1}{5040}(7)\cdots(1) + 1\right)\\
&= \dfrac{1}{8}(1 + 1) = \boxed{\dfrac{1}{4}}.\end{align}$$
