# Thread: Is this even possible?

1. ## Is this even possible?

Hey, I have this math problem that I don't think is possible...does anyone know if it is? (and if it is what is the answer?)

I have a Lot of markers in a drawer they are all red, black or purple. All but two are black, All but two are red, All but two are purple. How many markers in the drawer

The sad thing is this is my younger sister's homework...and no one in my house gets it. Thanks if you can help.

2. Originally Posted by VampireScooby
Hey, I have this math problem that I don't think is possible...does anyone know if it is? (and if it is what is the answer?)

I have a Lot of markers in a drawer they are all red, black or purple. All but two are black, All but two are red, All but two are purple. How many markers in the drawer

The sad thing is this is my younger sister's homework...and no one in my house gets it. Thanks if you can help.
You've got one of each colour... i.e. 3 markers

3. Originally Posted by VampireScooby
Hey, I have this math problem that I don't think is possible...does anyone know if it is? (and if it is what is the answer?)

I have a Lot of markers in a drawer they are all red, black or purple. All but two are black, All but two are red, All but two are purple. How many markers in the drawer

The sad thing is this is my younger sister's homework...and no one in my house gets it. Thanks if you can help.
You have 2 markers of each colour. There are total six markers.

4. Originally Posted by Shyam
You have 2 markers of each colour. There are total six markers.
Wrong. If there are two of each colour, then there are 2 reds, 2 blacks and 2 purples.

If there are 2 reds, there are 4 of another colour. So NOT all but 2 are red.

Same argument for blacks and purples.

HOWEVER...

If you have ONE of each colour, then there is 1 red, 1 black and 1 purple - making 3 markers.

1 red means all but 2 are red.
1 black means all but 2 are black.
1 purple means all but 2 are purple.

5. Let's say we have $T$ number of pens, and say $R,B,P$ denote the number of red, black and purple pens respectively.
Then, we know that $R+B+P=T$.
By the given properties, we know that
$T-2=R\qquad(1)$
$T-2=B\qquad(2)$
and
$T-2=P\qquad(3)$
hold.
By summing (1)--(3) side by side, we get
$3T-6=R+B+P$
..._...... $=T,$
which gives
$2T=6$
or simply
$T=3$.
Therefore, we have $3$ pens in total.