Multiples of six in an interval How many multiples of $6$ are there between the following pairs of numbers?
$0$ and $100$ and $-6$ and $34$.

*

*$16$ and $6$

*$17$ and $6$

*$17$ and $7$

*$16$ and $7$

My attempt:
Total number of integer between $0$ to $100 = 101$
So, total number of multiplier between $0$ to $100 = \left\lfloor\frac{101}{6}\right\rfloor = 16$
And, Total number of integer between $-6$ to $34 = 41$
So, total number of multiplier between $-6$ to $34 = \left\lfloor\frac{34}{6}\right\rfloor = 6$

But, somewhere it explained as:

Number of multiples of $6$ between $1$ and $100 = 100/6 = 16$
Since the range starts from zero, we need to take zero too. [zero is a multiple of every integer (except zero itself)].
So, answer $= 16+1 = 17$
Number of multiples of $6$ between $1$ and $34 = 34/6 = 5$
Since the range is $-6$ to $34$, we need to take $-6$ and zero.
So, answer $= 5+2 = 7$.

However, this small question and we can count on fingers.
But, what is wrong with my approach, can you explain it, please?

There are no multiples of $m$ bewteen $km+1$ and $km + (m-1)$ and the only multiple between $km$ and $km + (m-1)$ is $km$.
So....
The  multiples of $m$ between $km + 1$ and $jm$ will be the $lm$ for $k < l \le j$ or $0 < l-j \le j-k$ and there will be $j-k$ of them.
The multiples of $m$ between $km$ and $jm$ will be the $lm$ for $k \le l \le j$ or $0 \le l-j \le j-k$ and there will be $(j-k) + 1$.
The multiples of $m$ between $km + i;0\le i < m$ and $jm + n; n\le 0 < m$ will be $j-k$ if $i > 0$ and $(j-k)+1$ if $i= 0$.
So....
Between:
$0=6*0$ and $100=96+4$ will have $\frac {96}6 + 1$.
$−6=-6$ and $34=30+4$ will be $\frac {30-(-6)}6 + 1$.
$16= 12 + 4$ and  $6$ will be $\frac{12-6}{6}+1$
$ 6= 6$ and $17=12+5$ will be $\frac{12-6}{6} + 1$
Those all had the lower term itself being a multiple of $6$.
$7 = 6+1$ and $17= 12 + 5$ will be $\frac{12-6}{6}$.  This was the only one were the lowest term was not itself a multiple of $6$.
