Is there a trick to finding the number of odd numbers b/w two values? I know you could find the number of even numbers (since they are a multiple of two). For example the number of even numbers between $11$ and $30$ will be
$$n= \frac{28-12}{2} + 1 = 9  $$
I wanted to know is there a similar way to find the number of odd numbers b/w two extremes?
 A: You can use the exact formula. What is number of odd numbers between $11$ and $30$?
$$n = (29-13)/2 + 1 = 16/2 + 1 = 9$$
Let's list them out to make sure $: 13, 15, 17, 19, 21, 23, 25, 27, 29.$ There are $9$ of them, so that is correct.
By the way, your formula was calculated wrong. $$n = (28-12)/2 + 1 = 16/2 + 1 = 8 + 1 = 9 (12, 14,16,18,20,22,24,26,28)$$ 
A: Method 1:
If you can find the number of numbers, and you can find the number of even numbers, ....
Method 2:
The number of odd numbers between 11 and 30 is the same as the number of even numbers between 12 and 31.
A: If you use the formula $(b-a)/2 + 1$, then for some case it would give wrong answer.
Say, $a = 2$ and $b = 4$, so there is only one ODD nmumber between 2 and 4 (inclusive), and that is 3.
By using the formula, we get $(4-2)/2 + 1 = 2$, that is wrong; if we consider a small alter here, we will calculate the difference like this:
$Diff = (b - a) /2 + X$
Here $X$ will be either 0 or 1 depending on whether any one of the variables (a or b) is odd or not. If $a \mod 2 = 1 \lor b \mod 2 =1$, then $X$ will be 1, otherwise 0.
That should give correct answer.
A: use the A.P formula for finding the number of terms...say for no. of odd numbers between 1 and 11.... nth term= first term+(n-1)(common difference)......(*),where n=no. of terms,which we have set out to find in this case.
from (*) we have n= {(nth term-first term)/common difference}+1.....
For our problem, n= {(11-1)}/2}+1
                  =6
here common difference is 2 as an odd number occurs by adding 2 to the previous odd number etc.,
A: You can use this formula which gives you the number of integers congruent to ${n}\pmod p$ in the interval $a$ inclusive and $b$ inclusive :
$S =\lfloor\frac{n-a}{p}\rfloor+\lfloor\frac{b-n}{p}\rfloor+1$
So here, we get : $\lfloor\frac{1-11}{2}\rfloor+\lfloor\frac{30-1}{2}\rfloor+1 = \lfloor-\frac{10}{2}\rfloor+\lfloor\frac{29}{2}\rfloor+1=-5+14+1=10$
There is $10$ odd numbers between $11$ and $30$. If we disregard $11$, then there are $9$.
A: To get count of odd or even numbers between a range, follow the process as below:
Correct the Range to start and end with inclusive numbers as per question and then use following formula : 
(m - n)/2 + 1 where m is greater than n
Example:
All Odd numbers between 21 - 61
correct the range to make it inclusive of the numbers which make the range to 23 - 59
use the formula: 
(59 - 23)/2 + 1 => 19
All even numbers between 21 - 61
correct the range to make it inclusive of the numbers which make the range to 22 - 60
use the formula: 
(60 - 22)/2 + 1 => 20
A: For Calculating Even and Odd where S= starting number and E= ending Number
No. Of even = m.floor((e-s)/2)+(-(e%2)+1)
No. Of Odd = m.ceil((e-s)/2)+(e%2)
A: The Problem What is the number of EVEN or ODD integers between two numbers n and m? 
(where m>n)
Solution:


*

*Recall the formula (m-n+1)/2 

*Calculate it. 

*See below for how to interpret:


*

*If m&n are mixed signs*, you're done. That's your answer.

*If m&n are the same as the target**, round up.

*If m&n are different than the target, round down.



Details:


*

**mixed signs = one is even and the other is odd.

***"same as the target = if you're looking for evens, then whether m&n are both even. If you're looking for odds, then whether m&n are both odd.

