The Problem: Let f(x) be a polynomial of degree 10 with leading coefficient 1. If

f(1) = 1, f(2) = 2, f(3) = 3, . . . , f(10) = 10,

then what is f(11)?

Try to figure it out before reading on!

Applying our ten known values of f(x), we have ten linear equations:

Yuck. Have fun solving that. Let's try finding the roots of f(x) instead. We first write f(x) in terms of its roots:

Applying our ten known values (f(1) = 1, f(2) = 2, etc.) this time yields:

Hurm. This may be even messier than our first attempt; finding the roots of f(x) with these equations isn't going to be easy. We'll have to look for something different. We observe that each of the ten given values of f(x) is simply related to the x that yields it. Namely,

f(k) = k for each integer k from 1 to 10.

If we rewrite this as

f(k) - k = 0 for each integer k from 1 to 10,

we see that while we weren't able to find any roots of f(x), we know all the roots of f(x) - x. Specifically, if we let

g(x) = f(x) - x

then we know that x = 1, 2, 3, ... , 10 are roots of g(x) since g(x) = f(x) - x = 0 at each of these points. We know these are the only roots of g(x) because we are given that f(x) is a degree 10 polynomial, from which we deduce g(x) is also. Hence, we can write

f(x) - x = g(x) = (x - 1)(x - 2)(x - 3)...(x - 10),

so

f(x) = (x - 1)(x - 2)(x - 3)...(x - 10) + x.

Verify that this function satisfies f(1) = 1, f(2) = 2, f(3) = 3, etc. Now we can find f(11):

f(11) = (11 - 1)(11 - 2)(11 - 3)...(11 - 10) + 11 = 10! + 11.

If you think you've got this down, try the problems below. They're in order of increasing difficulty, and the buttons below will provide answers and hints.

A) Let f(x) be a polynomial of degree 8 with leading coefficient 1. Find f(9) if

f(1) = f(2) = f(3) = f(4) = f(5) = f(6) = f(7) = f(8) = 1.

B) Let f(x) be a polynomial of degree 8 with leading coefficient 1. Find f(9) if

f(1) = 1, f(2) = 4, f(3) = 9, . . . , f(8) = 64.

C) (Hard) Let f(x) be a polynomial of degree 7. Find f(9) if

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, f(8) = 1/8.

D) (From the 1984 USAMO - really hard!) P(x) is a polynomial of degree 3n such that

P(0) = P(3) = ... = P(3n) = 2
P(1) = P(4) = ... = P(3n - 2) = 1
P(2) = P(5) = ... = P(3n - 1) = 0
P(3n + 1) = 730.

Determine n.