Simplify $\frac{1}{n}\sum_{d\mid u} \varphi(d)\cdot 2^{\frac{nr}{d}}$

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I have been trying to solve the following problem:

Let $n$ and $r$ be positive integers. How many subsets of $\lbrace 1,2,\dots, nr\rbrace$ are there whose elements have a sum divisible by $n$?

Now, I have been making some calculations and arrived at the following expression, where $u$ is the largest odd divisor of $n$ and $\varphi$ is Euler's totient function:

$$\frac{1}{n}\sum\limits_{d\mid u} \varphi(d)\cdot 2^{\frac{nr}{d}}$$

Does anyone have an idea how to simplify this any further? Is that even possible?