Satvik's Polynomial

One thing that I've learnt when you have a polynomial such that the input is some constant plus or times the output is to create another polynomial. (This is particularly difficult to see if this is the first time you see a problem like this.)

In this case, let's construct a polynomial $P(x)$ that will exhibit the properties of the first five given values (namely $f(1)$ through $f(5)$ ). We will say that $P(x) = f(x) - (x+1)$ , so that 1, 2, 3, 4, and 5 are all zeroes of $P(x)$ . This yields $P(x) = a(x-1)(x-2)(x-3)(x-4)(x-5)$ , where $a$ is some constant (as $P(x)$ is not necessarily monic, which will become obvious shortly).

This is particularly helpful, as we can now say that $f(x) = P(x) + (x+1) = a(x-1)(x-2)(x-3)(x-4)(x-5) + (x+1)$ .

Now, let us evaluate $a$ . Finding $f(8)$ should do this, since we know that $f(8) = 7$ . We have $f(8) = 7 = a(7)(6)(5)(4)(3) + 9$ , which yields $a = \frac{-1}{1260}$ .

We can now find $f(9)$ fairly quickly. Note that $f(9) = a(8)(7)(6)(5)(4) + 10$ , which yields $f(9) = \frac{14}{3}$ after substituting $a$ . Our answer is thus $\boxed{17}$ .