Counting numbers in a sequence - explain "Add $1$ before you're done" rule I'm studying for the GRE, and my study book uses a rule that it never justifies for counting numbers in a sequence: "Add $1$ before you're done."
For example, how many multiples of $3$ are between $250$ and $350$? My study book says:
$$
348 - 252 = 96
$$
$$
\dfrac{96}{3} = 32
$$
Now "add one before you're done":
$$
32 + 1 = 33
$$
I follow the first few steps. Start and end with $252$ and $348$ because they are the first and last multiples of $3$ within the range, respectively. We divide by $3$ to count only the multiples of $3.$ But why add $1$?
 A: Arrgh. This apparently was written by someone who thinks that mathematics is merely arbitrary rules to be memorized rather than something to be understood. 
It starts by finding the distance between two multiples of $3$, then divides that distance by $3.$ In doing this, you are not calculating multiples of $3$, you are calculating the number of gaps of size three between the multiples. But the multiples themselves are the "fenceposts" that create the gaps. And there has to be one more "fencepost" than gap, to form the ends. For example:


*

*$0,3$. Distance = $3 - 0 = 3$, number of gaps = $3/3 = 1$, number of multiples = $2 = 1 + 1$.

*$0,3,6$. Distance = $6 - 0 = 6$, number of gaps = $6/3 = 2$, number of multiples = $3 = 2 + 1$.

*$3, 6, 9, 12$. Distance = $12 - 3 = 9$, number of gaps = $9/3 = 3$, number of multiples = $4 = 3 + 1$.

A: Whenever you want to count the number of things, align it to the sequence $1, 2, \ldots$.
In the sequence $1,\ldots,n$, there are exactly $n$ items. This should be abundantly clear.
In the sequence $0,\ldots,n$, there are exactly $n+1$ items. This should be abundantly clear once you understand the first statement.
When you ask "How many numbers are there between 30 and 40?" and you answer "10" instinctively, you've done subtraction. Subtraction starts you at 0, not 1 -- if we subtract 30 from every number between 30 and 40, the sequence becomes $0, \ldots, 10$.
Instead of subtracting the first number, I always subtract the number that gets me to 1 -- here, 29 (i.e., 30-29=1). And 40-29 is 11. Or, if you prefer, just add 1 to 10, in parallel to the $0,\ldots,10$ argument.
A: You can use the idea of Arithmetic Progression.

$$t_n=a+(n-1)d$$ in any arithmetic progression where $t_n$ is the $n-th$ term which readily implies that $n\in\mathbb Z$.

For your first example. Between $250$ and $350$ the first multiple of $3$ is $252$ and the last multiple is $348$ and this forms an AP. So, we can write-$$348=252+(n-1)\times3$$ Solving you readily get $n=33$.
A: There are just as many multiples of $3$ between $252$ and $348$ (inclusive) as there are between $0$ and $348-252$, that is, $0$ and $96$.
Now what one does is a matter of taste. 
The number between $0$ and $96$ is the same as the number from $3$ to $99$, which is the same as the number of integers between $1$ and $33$.
Or else there are $32$ between $3$ and $96$. But don't forget about $0$. 
Or else there are just as many as there are integers from $0$ to $32$. Again, don't forget about $0$. There are $33$ of these. 
Remark: I prefer to figure it out each time. Rules are hard to remember. But I am lucky. I don't have to solve too many too easy questions in too little time.
