Help with a riddle please

[MooMoo_Cowfreak]

So I'm playing Neverwinter Nights 2, and there's this riddle. It's answer should be a three digit number, but I can't seem to find it. (I have tried gamefaqs but the puzzle is randomized so there is no definite answer,)

9 are we when taken all 3.

Last one slain than six remain.

None survive, our first less five.

I've tried but I can't seem to find it.

Thanks!

[bballa48]

513

[Zageron]

Answer = xyz in one number no multiplication.
x + y + z = 9
x + y - z = 6
x = -5

Try that. :S

[phe0nixblade]

the answer is va-gina

[MooMoo_Cowfreak]

Originally Posted by Zageron

Answer = xyz in one number no multiplication.
x + y + z = 9
x + y - z = 6
x = -5

Try that. :S

It doesn't do negatives sorry.

Originally Posted by bballa48

513

I'll try it along with a lot of other answers when I get them.

[Zageron]

Originally Posted by MooMoo_Cowfreak

It doesn't do negatives sorry.

Your supposed to use that formula / system of equations to figure it out. Fine I'll do it.

Edit: It was wrong anyways, I think bballa48 is right.

[MooMoo_Cowfreak]

Originally Posted by Zageron

Your supposed to use that formula / system of equations to figure it out. Fine I'll do it.

oh i see.

EDIT: I'm trying to solve it but I am stuck. Ugh.

EDIT2: Oh well it turns out it's unsolvable. If x is = to -5 you still get a negative number in there.

[abro13]

I can only come up with 423 ,333 ,243 ,153 , or 063. You can use zeros, right?

[TheCookieMonster]

6 (9) 5 (8) 0 (3) [4]

650

983

984

etc.

I don't know if I'm right or wrong...

[MrRubix]

a+b+c=9
a+b=6
a-5=0

a=5, so b=1 and c=3

try 513

[Zageron]

Originally Posted by MrRubix

a+b+c=9
a+b=6
a-5=0

a=5, so b=1 and c=3

try 513

This is what my equation was aiming for, but I messed it up D: