When you take away two

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

• Correct, I guess I made this way too easy Jun 26 '20 at 14:38
• Not necessarily, I think it's just that, on this site, a few of us are attuned to this kind of trickery. Jun 26 '20 at 14:40

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...)

• Great answer! There is still yet another answer, since this is not the alternative answer I was thinking of. Jun 26 '20 at 15:20

5

So,

When you take away two letters from Five
Five - Fe = IV (4 in roman numerals)

• Correct, this is the alternate answer I was thinking of Jun 27 '20 at 17:34

$$2^0$$

Explanation:

$$2^0=1$$. But if you take away the $$2$$, you're left with $$0$$.

• This is clever. I didn't even consider equations as possible solutions Jun 27 '20 at 22:41

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.

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

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.

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.

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

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.