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?
Best Answer
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.