An algebra problem by Wildan Bagus Wicaksono

Since $f(1)=0$ , $f(2)=0$ and $f(3)=0$ , we can write given polynomial in form of $f(x)=a(x-1)(x-2)(x-3)(x-\alpha_{4})$ , where $\alpha_{4}$ is its fourth root and $a$ is a real coefficient.

Then, we have $f(4)=a(4-1)(4-2)(4-3)(4-\alpha_{4})=6$

$\Rightarrow 6a(4-\alpha _{4})=6$

$\Rightarrow a(4-\alpha _{4})=1$

$\Rightarrow a=\frac{1}{4-\alpha _{4}}$

Next, we have $f(5)=a(5-1)(5-2)(5-3)(5-\alpha_{4})=72$

$\Rightarrow 24a(5-\alpha _{4})=72$

$\Rightarrow a(5-\alpha _{4})=3$

$\Rightarrow \frac{1}{4-\alpha _{4}}(5-\alpha _{4})=3$

$\Rightarrow 5-\alpha _{4}=12-3\alpha _{4}$

$\Rightarrow 2\alpha _{4}=7\Rightarrow \alpha _{4}=\frac{7}{2}$ .

Thus, $a=\frac{1}{4-\frac{7}{2}}=2$ .

Now, we have general expression for this polynomial $f(x)=2(x-1)(x-2)(x-3)(x-\frac{7}{2})$ .

$f(2010)=2(2010-1)(2010-2)(2010-3)(2010-\frac{7}{2})$

$f(2010)=2\times 2009\times 2008\times 2007\times \frac{4013}{2}=2009\times 2008\times 2007\times 4013$

Last digit of this product will be the last digit of $9\times 8\times 7\times 3$ , which is $\boxed{2}$ .

Let $f(x) = ax^4+bx^3+cx^2+dx+e$ . Then we have:

$\begin{aligned} \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 16 & 8 & 4 & 2 & 1 \\ 81 & 27 & 9 & 3 & 1 \\ 256 & 64 & 16 & 4 & 1 \\ 625 & 125 & 25 & 5 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ e \end{bmatrix} & = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 6 \\ 72 \end{bmatrix} \end{aligned}$

Since we need to find the last digit of $f(2010)$ , all terms with $x$ end with 0 and we need only to find $e$ . We can do that using Cramer's rule as follows:

Let $A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 16 & 8 & 4 & 2 & 1 \\ 81 & 27 & 9 & 3 & 1 \\ 256 & 64 & 16 & 4 & 1 \\ 625 & 125 & 25 & 5 & 1 \end{bmatrix} \implies \det (A) = 288$

Let $A_5 = \begin{bmatrix} 1 & 1 & 1 & 1 & 0 \\ 16 & 8 & 4 & 2 & 0 \\ 81 & 27 & 9 & 3 & 0 \\ 256 & 64 & 16 & 4 & 6 \\ 625 & 125 & 25 & 5 & 72 \end{bmatrix} \implies \det (A_5) = 12096$

Therefore, $e = \dfrac {\det(A_5)}{\det(A)} = \dfrac {12096}{288} = 42$ . $\implies$ the last digit of $f(2010)$ is $\boxed{2}$ .