How many multiples of 3 are between 10 and 100? (SAT math question) 
In the figure above, circular region A represents all integers from 10 to 100, inclusive; circular region B represents all integers that are multiples of 3; and circular region C represents all squares of integers. How many numbers are represented by the shaded region?


*

*a) 24 

*b) 25 

*c) 26

*d) 27

*e) 28


Here's my train of thought:
All we need to know is this set of numbers is from 10-100 (which I'm assuming does include 10 and 100?) and is multiples of 3. Instead of listing out every multiple of 3, I can have the largest multiple of 3 less than or equal to 100, which is 99, subtract the smallest multiple of 3 more than or equal to 10, which is 12.
99 - 12 = 87
Then to find the number of times 3 fits into 87, divide 87 by 3.
87 / 3 = 29
Agh! 29 is not one of the answer options provided! 
Assuming what I've done so far is correct, now I'm thinking that for some reason 1 has to be subtracted from 29 
29 - 1 = 28
to get 28, which is one of the options provided and also is the correct answer. 
But why would I need to subtract 1 from 29? (Or is that not the right way to find the answer?)
 A: $$\left\lfloor\frac{100}3\right\rfloor-\left\lfloor\frac{10}3\right\rfloor$$

Alternatively, 
the first term is $12,$ the last being $99$ forming an Arithmetic Series with Common Difference $=3$
So, the $n$th term $=99=12+(n-1)3$
A: There are $30$ multiples of three between $10$ and $100$ (and it doesn't matter whether we include $10$ and $100$).  You are correct that they range from $12$ through $99$, but you made a fencepost error by not adding one.  So there are $30$ numbers in the shaded part plus the center of the diagram.  The center has numbers that are between $10$ and $100$, squares, and multiples of $3$ (and hence $9$ as they are squares).  These are $36$ and $81$.  Subtracting those two gets you to $28$.
A: There are $33$ (positive) multiples of $3$ less than $100$, but we need to throw out $3$, $6$, and $9$, so that takes us down to $30$.  However, the problem also wants us to throw out any squares that are multiples of $3$, which means square multiples of $9$, which means $9$, $36$, and $81$.  We've already thrown out $9$, so we need only throw out the other two, leaving the count at $28$.
