For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$ For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following expression:
$$\binom{5^{m-k}}{2}$$ which is not only too good to be true but it turns out it is wrong. Any suggestions on how I should approach this?
Edit: After reading through @heropup's example, I am starting to realize that I may have forgotten $5^k$ term and so the following could be correct.
$$5^k \binom{5^{m-k}}{2}$$
Does that sound correct?

Your original approach of doing casework on intervals works. The key (at least to my method) is to interpret $$\left\lfloor\frac{a}{b}\right\rfloor$$ as the number of positive multiples of $b$ that are less than or equal to $a,$ where $a$ and $b$ are positive integers. Your list of disjoint intervals that cover all of the $l$ is $$[1,5^k -1],[5^k, 2\cdot 5^k -1],[2\cdot 5^k,3\cdot 5^k -1],\ldots, [5^{m-k-1}\cdot 5^k,5^{m-k}\cdot 5^k -1].$$
In the first interval, none of the elements have a positive multiple of $5^k$ less than or equal to the element. For each integer in the $t^{\text{th}}$ interval for $t\ge 2$ the number of positive multiples of $5^k$ that are less than or equal to the element is $t-1.$ There are $5^k$ elements in each interval after first interval. So the answer is
\begin{align*}
\sum_{n=1}^{5^{m-k}-1}{5^k\cdot n} &= 5^k\cdot\sum_{n=1}^{5^{m-k}-1}{n}\\
&= 5^k \cdot \frac{(5^{m-k}-1)5^{m-k}}{2}\\
&= \frac{5^{m}\cdot (5^{m-k}-1)}{2}.
\end{align*}
This is the same as $5^k \cdot \binom{5^{m-k}}{2}.$
