Calculating a probability mass function (sufficient statistic) 
Let $X_1,\ldots,X_n$ be independent random variables with densities:
$$f_{X_k}(x\mid\theta)= \begin{cases}  e^{k\theta-x}& x\geq
k\theta\\   0& \text{otherwise}  \end{cases} $$
Find the pmf for $T=\min_{k}\left(\frac{X_k}{k}\right)$.

So given the pdf of $X_k$, the pdf of $Y_k=\frac{X_k}{k}$ is given by 
$$f_{Y_k}(y\mid\theta)=\begin{cases}
 ke^{k(\theta-x)}& x\geq \theta\\ 
 0& \text{otherwise} 
\end{cases}
$$
Thus for each $k$, $1\leq k\leq n$, I need to find $\text{P}(T=Y_k).$  This is the same as calculating the probability that the remaining random variables are all greater than $Y_k$ for some fixed $Y_k=y_k$, and then integrating over all possible values of $Y_k$ weighted according to its pdf, where by independence, 
$$\text{P}(Y_1,\dots,Y_{k-1},Y_{k+1},\dots,Y_n > y_k)=\prod_{j=1,\dots,k-1,k+1,\dots,n}\int_{y_k}^{\infty}je^{j(\theta-y)} \, dy=\prod_{j=1,\dots,k-1,k+1,\dots,n}e^{j(\theta-y_k)}.$$
Thus we have:
$$\text{P}(T=Y_k)=\int_{\theta}^{\infty}\Big(\prod_{j=1,\dots,k-1,k+1,\dots,n}e^{j(\theta-y_k)}\Big)ke^{k(\theta-y_k)} \, dy_k = k\int_{\theta}^\infty e^{\frac{n(n+1)}{2}(\theta-y_k)} \, dy_k=\frac{2k}{n(n+1)}.$$
Whew!  Ok this sums to $1$ as $k$ ranges between $1$ and $n$, which gives me confidence I did this correctly.  But on the other hand, it's now requested that I show that $T$ is a sufficient statistic for the parameter $\theta$.  Which by a theorem in the book means that the pdf of the joint distribution $(X_1,...,X_n)$, divided by the pmf of $T$, should be a function which is independent of $\theta$.  
But the pmf of $T$ is itself independent of $\theta$, that would mean that the pdf of $(X_1,\dots,X_n)$ would also have to be independent of $\theta$, which doesn't appear to be true, since by independence it's just the product of the individual density functions and $\theta$ does not disappear upon taking that product.
The book uses a different and much easier theorem to prove that $T$ is a sufficient statistic, so I'm not looking for a proof of that, but instead would like to find out what's wrong with my method.  Thanks.
 A: Sorry, I missed something and got my last answer mangled.  I've deleted that.
Notice that
$$
\Pr(Y_k>x) = \int_x^\infty ke^{k(\theta-w)} \,dw = e^{k(\theta-x)}.
$$
So
\begin{align}
\Pr(\min>x) & = \Pr(\text{for all $k$, }Y_k>x) \\[10pt]
& = e^{\theta-x} e^{2(\theta-x)} e^{3(\theta-x)}\cdots e^{n(\theta-x)} \\[10pt]
& = e^{n(n+1)(\theta-x)/2}.
\end{align}
Consequently for $x>\theta$, we have
$$
f_\min(x) = \dfrac{d}{dx}(1- e^{n(n+1)(\theta-x)/2}) = \frac{n(n+1)}2\cdot e^{n(n+1)(\theta-x)/2}.
$$
Dividing by $e^{\theta-x} e^{2\theta-x} e^{3\theta-x}\cdots e^{n\theta-x}$ $= e^{n(n+1)\theta/2 - nx}$, one sees that the $\theta$ cancels and the $x$ doesn't.
Here's what seems to be your error: You wrote:

But the pmf of $T$ is itself independent of $\theta$

That's not true.  The pdf is derived above.  What you found was not the pdf of $T$, but the pmf of the discrete random variable whose value is the index $k$ for which $T=Y_k$.  The random variable $T$ itself is a continuous random variable.  Maybe you're confusing $T$ with that discrete random variable.
