This might sound trivial, but I am wondering why, for example if you do:
$\lfloor10/8\rfloor = 1 \implies 1$ number $[1,10]$ divisible by $8$: $\{8\}$
$\lfloor10/3\rfloor = 3 \implies 3$ numbers $[1,10]$ divisible by $3$: $\{3,6,9\}$
$\lfloor23/4\rfloor = 5 \implies 5$ numbers $[1,23]$ divisible by $4$: $\{4,8,12,16,20\}$
$\ldots$ etc.
Why is this true?
Because of division. What you're essentially doing when you do:
$$10\over 3$$
Is asking, how many times does $3$ go into $10$? The answer is $3.\bar3$, but the leading whole $3$ tells you that there were $3$ whole times plus some fractional part. Since $3$ went into $10$ three whole times, there must be $3$ numbers smaller than or equal to $10$ that are multiples of $3$. Flooring the division just gives you the whole part.
Let's make it general. If we are seeking the number of natural numbers divisible by say $y$ up to a number say $x$, then $\lfloor x/y\rfloor$ is the number that we are seeking because those numbers are $1y,\ 2y,\ 3y, ..., ay\ $ where $ay$ is the biggest number divisible by $y$ with $ay \le x$. Then notice that
If $ay = x$, we directly have $a = \frac{x}{y} = \lfloor x/y\rfloor$ so we have $a$ many such numbers (from $1y$ to $ay$).
If $ay < x$, we have $x-ay = k$ with $k < y$, implies $a = \frac{x-k}{y}$ so $a = \lfloor x/y\rfloor$ again because we know that $k < y$. So again, we have $a$ many such numbers (again from $1y$ to $ay$).
Using the division lemma, the division $b/a$ gives us a sum
$$b = qa + r$$
with $0 \le r < a$. Now there are exactly $q$ multiples of $a$ less than or equal to $b$; namely, they are $a, 2a, 3a, \dots, qa$. Furthermore,
$$\lfloor \frac b a \rfloor = \lfloor q + \frac r a\rfloor = q$$ since $0 \le r/a < 1$.
Assume that $x$ and $y$ are positive integers. Then
$ 1y < 2y < \cdots < ny \le x < (n+1)y $ if and only if $\left \lfloor \dfrac xy \right \rfloor = n$
