Consider all polynomials f(x) with integer coefficients such that f(200)=f(7)=2007 and 0<f(0)<2007.?

Question

Show that the value of f(0) does not depend on the choice of polynomial and find f(0).

Answer 1

Quick question: should that read "...such that f(2000)...." instead of "....such that f(200)...."?

NOTE: the answer below mine assumes that you are talking about a quadratic polynomial. The questions specifically states "all polynomials" (ie, not just x^2....there is nothing stopping us from starting with x^2000000). Therefore it is not a general solution.

Answer 2

Here is how I approached the problem. If f(x) = (x-7)(x-200) + 2007, then f(7)=2007, f(200) = 2007, but f(0) = 3607, which does not satisfy the specifications of the function. However, if we make f(x) = - (x-7)(x-200) + 2007, f(0) = 607 and it is still true that f(7)=2007 and f(200) = 2007. Next, suppose that the function is cubic and f(x) = (x - 7)(x - 200)(x - a) + 2007. If you simplify this expression, then calculate the values of a that allow f(0) to be between 0 and 2007, you will discover that a = 1 is the only solution and f(0) turns out to be 607 again.

Answer 3

Let f(x) = a_nx^n + ... + a_0.
Since f(7) = 2007, we have a_0 = f(0) = 5(mod 7)
Further f(200) = 2007 = 7(mod 200)., so a_0 = 7(mod 200)
Now since 7 and 200 are relatively prime, the
Chinese remainder theorem tells us that the 2 congruences
above have a unique solution mod 200*7 = 1400.
Let's find it.
We have a_0 = 5 + 7t
so 5 + 7t = 7(mod 200)
7t = 2(mod 200).
Solving, we find t = 86(mod 200)
and a_0.= 5 + 7(86 + 200u), say
Thus a_0 = 607(mod 1400).
Since 0 < a_0< 2007, a_0 = 607 is
the only possible solution.
Only problem is that this value doesn't
guarantee that f(200) = f(7) = 2007.
We may get different values of the polynomials,
depending on the coefficients. I'll look at that
part of the problem some more.

Answer 4

Let's write f(x) = xg(x) + c, where g(x) is another polynomial and c is a constant. Then f(0) = c.

So if we can calculate c from the information given, we're done.

Well, c = 2007 - 200*g(200)
Also, c = 2007 - 7*g(7)
So c is congruent to 2007 both mod (200) and mod (7).

We're being asked to exhibit a unique solution in the interval [1,2006].

Well, if you know the Chinese Remainder Theorem, you know how to find one solution R. And then when you find it, you know that the set of all solutions is just the set of all R + k * 1400, for integers k (Because 7 and 200 are relatively prime, and 7*200 = 1400.) So there are either exactly 1 or exactly 2 possible solutions in [1,2006].

I'll let you do the calculations to finish the problem from there. ;)