A marble puzzle

[jason_t]

A jar contains 2005 red marbles, 2005 blue marbles, and 2005 white marbles. You also have a stock of extra marbles. You repeatedly reach into the jar and draw three marbles.

1. If the three marbles are red, you put a red marble back, and add a white marble to the jar.
2. If the three marbles are blue, you add two red marbles to the jar and remove a white marble.
3. If the three marbles are white, you put one white marble back and add a blue marble.
4. If two marbles are red and one is blue, you put two blue marbles in the jar and remove two more red marbles.
5. If two marbles are red and one is white, you put the red marbles back.
6. If two marbles are blue, and one is red, you take another red marble out of the jar and add three white marbles to the jar.
7. If two marbles are blue, and one is white, you put a blue marble back, and the white one back.
8. If two marbles are white, and one is red, you put the red one back, and add a blue marble to the jar.
9. If two marbles are white, and one is blue, you put both whites one back.
10. If one marble is red, one is blue and the third is white, you put the red marble back.

Assume that at some point the operations you perform result in a single marble remaining. Can you determine its color?

Answer revealed in six hours.

[IggyWH]

That's easy... it's white.

[TStoneMBD]

[ QUOTE ]
That's easy... it's white.

[/ QUOTE ]

sounds good to me.

[private joker]

[IggyWH]

[ QUOTE ]


[/ QUOTE ]

You know, there's surgery to fix that problem you have...

Don't believe bitches when they say size doesn't matter.

[gumpzilla]

You specify that if I draw two blues and a red, I need to draw another red from the jar. Doesn't this assume that there are more reds in the jar? What if there aren't?

I suspect this is irrelevant, as I think I get the point here, but it's problematic anyway.

EDIT: I'm pretty sure I have the solution, which I'll PM you, but I really think you need to fix some of these procedures.

[jason_t]

[ QUOTE ]

You specify that if I draw two blues and a red, I need to draw another red from the jar. Doesn't this assume that there are more reds in the jar? What if there aren't?

I suspect this is irrelevant, as I think I get the point here, but it's problematic anyway.

[/ QUOTE ]

The statement of the question stipulates that some how, some way you reach a point where only one marble remains and never encountered an impossible step.

[ QUOTE ]
Assume that at some point the operations you perform result in a single marble remaining. Can you determine its color?

[/ QUOTE ]

[jason_t]

Bizzle. Surprised at the lack of attempts.

[benthehen]

I may have made several mistakes. I got that the color can be determined and it must be red. Furthermore, it can only come from the 10th operation. I won't write out the entire argument, since its kinda long. Instead I'll give an outline of the argument:

Suppose the last color is blue. Suppose before the last operation we have a reds, b blues, and c whites. Suppose the last operation adds x reds, y blues, and z whites (note that x,y,z can be negative.) So for the last color to be blue we must have a+x = 0, b+y = 1, and c+z=0. Thus we have a = -x, b = 1-y, and c = -z. Now, a,b, and c must all be greater than or equal to zero. So we can eliminate some operations. For each of those operations that remain, we find a,b, and c such that after that operation is performed we have one blue remaining. However, once we find such a,b, and c for a given operation we can show that we do not have enough reds, blues, or whites to perform that operation. Hence, we arrive at a contradiction for each operation. Example:

Consider rule #3. This rule adds one blue and subtracts two whites. So if we have zero reds, zero blues, and two whites, then after performing this operation we will have one blue. However, we could never perform this operation because we need at least three whites to do so (and we only started off with two.)

Using this argument we will arrive at contradictions for blue and white, but not for red. If we have one red, one blue, and one white, then 10th operation is sure to follow. Giving us one red remaining.

[LethalRose]

easiest way to solve this is just to write a program to solve it, create a string and write some if statements.