Something not so easy

Let $g(x) = xf(x)-1$ .

Since $f(x)$ is a 8th-degree polynomial, the degree of $g(x)$ is 9.

Note that $rf(r)-1 = 0$ , for $r = 1,2,3,...,8,9$ .

That is, $x = 1,2,3,...,8,9$ are the roots of $g(x)$ .

We can write $g(x)$ as follows:

$g(x) = A(x-1)(x-2)(x-3)...(x-8)(x-9) = xf(x)-1$ , where $A$ is a real number.

If $x = 0$ , we have $g(0) = A(-1)(-2)(-3)...(-8)(-9) = 0f(0)-1$ .

That is, $A = \dfrac{1}{9!}$ .

If $x = 10$ , we have $g(10) = A(10-1)(10-2)(10-3)...(10-8)(10-9) = 10f(10)-1$ .

$A(9!) = 10f(10)-1$

$\{\dfrac{1}{9!}\}(9!) = 10f(10)-1$

$10f(10) = 2$

Therefore, $\dfrac{1}{f(10)} = \dfrac{10}{2} = \boxed{5}$ .

Consider the product for positive integer $r$ :

$\displaystyle P_r = \prod_{k=1, k \ne r}^9 (x-k) = \begin{cases} 0 & \text{if }k \ne r \\ (-1)^{r-1}(r-1)!(9-r)! & \text{if }x=r \end{cases}$

Note that $P_r$ has a degree of 8. Therefore, $f(x)$ can be expressed as:

$\displaystyle f(x) = \sum_{r=1}^9 \frac {(-1)^{r-1}P_r}{r(r-1)!(9-r)!} = \sum_{r=1}^9 \frac {(-1)^{r-1}P_r}{r!(9-r)!} = \frac 1{9!} \sum_{r=1}^9 (-1)^{r-1}\binom 9r P_r$

For integer $x > 9$ , we can express $P_r = \dfrac {\prod_{k=1}^9 (x-k)}{x-r}$ and $f(x)$ as follows:

$\begin{aligned} f(x) & = \frac 1{9!} \sum_{r=1}^9 (-1)^{r-1}\binom 9r \frac {\prod_{k=1}^9 (x-k)}{x-r} \\ & = \frac 1{9!} \sum_{r=1}^9 \frac {(-1)^{r-1}\binom 9r(x-1)!}{(x-r)(x-10)!} \\ & = \frac {(x-1)!}{9!(x-10)!} \sum_{r=1}^9 \frac {(-1)^{r-1}\binom 9r}{x-r} \end{aligned}$

Then we have:

$\begin{aligned} f(10) & = \frac {9!}{9!(0!)} \sum_{r=1}^9 \frac {(-1)^{r-1}\binom 9r}{10-r} & \small \color{#3D99F6} \text{Since }\binom nr = \binom n{n-1} \\ & = \sum_{r=1}^9 \frac {(-1)^{r-1}\binom 9{9-r}}{1+9-r} \\ & = \sum_{r=0}^8 \frac {(-1)^{r-1}\binom 9r}{r+1} \\ & = \sum_{r=0}^9 \frac {(-1)^{r-1}\binom 9r}{r+1} + \frac 1{10} & \small \color{#3D99F6} \text{Note that } (1-t)^n = \sum_{k=0}^n \binom nk t^k \\ & = \int_0^1 (1-t)^9 dt + \frac 1{10} \\ & = - \frac {(1-t)^{10}}{10} \bigg|_0^1 + \frac 1{10} \\ & = \frac 1{10} + \frac 1{10} = \frac 15 \end{aligned}$

Therefore, $\dfrac 1{f(10)} = \boxed 5$ .

One simple method is to construct and extend a difference table.

Another approach is to explicitly construct the polynomial.

Note that $xf(x)-1$ is a polynomial of degree $9$ with roots at $x=1,2,\ldots,9$ . This means it must have the form $k(x-1)(x-2)\cdots (x-9)$ , where $k$ is a constant.

By comparing the constant terms, we can see that $k=\frac{1}{9!}$ . So $f(x)=\frac{1}{9!x} (9!+(x-1)(x-2)\cdots (x-9))$

We want to find $f(10)$ . This actually works out neatly, since $(10-1)(10-2)\cdots (10-9)$ is just $9!$ .

Substituting in, we find $f(10)=\frac{1}{5}$ so the required answer is $\boxed5$ .