I am an odd number.
When you take away two,
I become even.
What am I?
Edit: @hexomino got my original answer but I have come up with another valid answer.
Edit 2: @QuantumTwinkie got my other answer. It's really cool to see so many other creative solutions.
I think you are
Eleven
Explanation
When you take away the first two letters it becomes even
Alternatively, you are:
The number 9 (which is odd), on a seven-segment display.
If you 'take away two' - specifically:
Two segments - the one at the top and the one at the bottom - you end up with the number 4 (which is even):
This would also work with matchsticks... (and a similar technique could be used to change (e.g.) 19 into 14, 29 into 24, etc...)
How about
5
So,
When you take away two letters from Five
Five - Fe = IV (4 in roman numerals)
How about
$2^0$
Explanation:
$2^0=1$. But if you take away the $2$, you're left with $0$.
Another possibility similar to @Stiv's:
You are 7
Explanation:
In braille, the digit 7 is represented as ⠛. Removing the right two dots yields ⠃, the braille representation for 2. All other odd digits have at most two dots, so this is the unique single digit odd number with this property. Of course, any odd number ending in 7 will have this property too.
How about
12001 - 2 = 11000
when written as:
"twelve thousand one"
removing letters "t", "w", "o", re-ordering to
"eleven thousand"
Also works with certain other powers of 10: hundred, million, etc...
My guess is based on @Quantum Twinkie's answer. There are many Roman Numeral paths, for example:
XXI = 21 Take away X and I then X is even XXIX Take away X and I and so on
Here's another answer:
101
Explanation:
Written out in numerals, "One hundred and one" becomes "one hundred" when you remove the last two words: "and" and "one"
Admittedly not the most elegant solution, and there are similar solutions ad infinitum.
Odd Take away the last two letters, you get O, which is even.
My two cents. (Go on, take them.)
$\frac{36}{12}=3$ is odd.
Take away $2$:
$\frac{36}{1}=36$ is even.
How about
111 in binary, which is 7. Take away two right-most ones and you get 100, which is 4 (even).
There could be infinitely many solutions.
a) Pick any number in the following pattern:
[ANY NUMBER OF ANY LENGTH] [EVEN DIGIT] [ODD DIGIT]
Remove the last digit and any digit from the first set of numbers and the resulting number will be even.
b) Pick any number in the following pattern:
[ANY EVEN NUMBER OF ANY LENGTH] [TWO ODD DIGITS]
Remove the last 2 digits and the resulting number will be even.
What about simply
Uneven
Taking away two letters:
Even
An infinite amount of solutions, you are:
$\frac{4n+2}{2}, \text{where } n\in\mathbb{N}$
For example, if $n=3$, then $\frac{4(3)+2}{2} = 7$, which is odd
Taking away two:
Taking away any 2, you get $4n+2$ or $\frac{4n}{2}$ or even $4n$
These are always even
Another answer not yet mentioned:
You are number $-2^{53}+1$ stored in IEEE 754 double-precision format (or similar), e.g. C++'s
double
because
after subtracting 2, the result $-2^{53}-1$ cannot be represented exactly anymore (since the format is able to store only 53 significant binary digits), so it's rounded to nearest even integer. Try it online!
Another interpretation:
"Six", an odd number of characters, take away two is "four", an even number of characters.
I also think you are:
11
Explanation:
"an odd number" has 11 characters; "away two" has 7; "even" has 4. 11 - 7 = 4.
How about One. If you take away 2 letters 'ne', what remains is O which looks like 0. 0 is an even number
Admittedly pile-jumping, but how about:
೨೭, which is twenty-seven in Kannada. When you take away 2, you are left with ೨, which is two.
29 in binary is -11101. Removing two 1s( 3rd & 5th) , we get 110, an even number 6. Like that many such possibilities.
Potentially ALL odd numbers bigger than 1. If you "take away" 2 from the number line, all numbers above collapse down by one and all the numbers that used to be odd become even.
