polynomials

Well, we know from the remainder left by dividing by (x-a) = f(a). So you get a linear system.
Say our polynomial p(x) = $a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ . Putting in
dividing p(x) by (x-1), the remainder =1. so $f(1) = 1$
like wise $f(2) = 2, f(3) = 3, f(4) = 4, f(5)$ and the remainder left by divind x-6 = $f(6)$ . Lets call the remainder p(x) divided by x-6 = mSo we get
We get $f(1) = a_5 + a_4 + ... + a_0$
$f(2) = a_52^5 + a_42^4 + ... + a_0$ etc...
which is the linear system
$\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 2^5 & 2^4 & 2^3 & 2^2 & 2^1 & 1 \\ 3^5 & 3^4 & 3^3 & 3^2 & 3^1 & 1\\ 4^5 & 4^4 & 4^3 & 4^2 & 4^1 & 1 \\ 5^5 & 5^4 & 5^3 & 5^2 & 5^1 & 1 \\ 6^5 & 6^4 & 6^3 & 6^2 & 6^1 & 1 \end{bmatrix} \begin{bmatrix} a_5 \\ a_4 \\ a_3 \\ a_2 \\ a_1 \\ a_0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ m \end{bmatrix}$ . Solving this you get
$a_5 = -\frac{m-6}{120}$ since for a deg 5 polynomial, its coeff cant be 0, The solution is the remainder when p(x) is divided by x-6 is any number except 6.