SOLUTION. Clearly $f(n, x)$ is the number of combinations with repetition of $x+1$ elements taken from $n-x$ in $n-x$. That is,

$$f(n, x)=C R(x+1, n-x)=\left(\begin{array}{c} (x+1)+(n-x)-1 \\ n-x \end{array}\right)=\left(\begin{array}{l} n \\ x \end{array}\right)$$

We will prove that the $n$ sought are all of the form $p^{a}$ with $a$ a positive integer. Let $m_{p}$ be the $p$-part of the positive integer $m$, that is, if $m=p^{a} q$ (with $q \geq 1$ an integer), $m_{p}=p^{a}$, where $a \geq 1$ is an integer. Now we will prove the following preliminary result:

If $m_{p}=p^{a}$, then $(m-i)_{p}=i_{p}$ for each $i \in\left\{1,2, \ldots, p^{a}-1\right\}$.

Indeed, if $i_{p}=p^{k}$ then $k<a$ and it is obvious that $p^{k} \mid(m-i)$, so $i_{p} \leq(m-i)_{p}$. Conversely, if $(m-i)_{p}=p^{k}$, it must be that $k<a$ because otherwise $p^{a} \mid i$. Now, $p^{k} \mid i$ because $p^{k} \mid m$ and $p^{k} \mid(m-i)$. That is, $(m-i)_{p} \leq i_{p}$.

Next, we will prove that if $p$ is a prime and $n$ an integer greater than 1. Then $p$ divides $\left(\begin{array}{l}n \\ x\end{array}\right)$ for all $x \in\{1,2, \ldots, n-1\}$ if and only if $n=p^{a}$ with $a$ an integer.

If $p\left(\begin{array}{l}n \\ x\end{array}\right)$ for all $x \in\{1,2, \ldots, n-1\}, p\left(\begin{array}{l}n \\ 1\end{array}\right)=n$. Setting $n_{p}=p^{a}$, we have:

$$\left(\begin{array}{c} n \\ p^{a} \end{array}\right)=\frac{n(n-1) \ldots\left(n-p^{a}+1\right)}{p^{a}\left(p^{a}-1\right) \ldots 2 \cdot 1}$$

and by the preliminary result, we conclude that the $p$-part of $\left(\begin{array}{c}n \\ p^{a}\end{array}\right)$ is 1, so $n=p^{a}$.

Conversely, if $n=p^{a}$, for each $x \in\left\{1,2, \ldots, p^{a}-1\right\}$,

$$\left(\begin{array}{l} n \\ x \end{array}\right)=\frac{p^{a}\left(p^{a}-1\right) \ldots\left(p^{a}-x+1\right)}{x(x-1) \ldots 2 \cdot 1}$$

and again by the preliminary result, the $p$-part of $\left(\begin{array}{l}n \\ x\end{array}\right)$ is $\frac{p^{a}}{x_{p}}$, which is a multiple of $p$ because $x<p^{a}$.