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# Stacking the chips at Morty's

## Question

It was amateur night at Morty's. Alex had finished

his penultimate act of mysterious prestidigitation.

Preparing for the finale, he handed a pile of poker

chips to Jamie and, after turning his back to the

table, asked Jamie to construct three equal stacks

of at least four chips.

Back still to the table, Alex next asked Davey to

call out at random a number between 1 and 12.

Suppose Davey calls out the number x.

With the chips still out of his sight, and without

knowing the number of chips in each stack, Alex

finally gave instructions to Ian to shift chips among

the piles, in a manner that left x chips in the center

stack.

What is the method?

To clarify, neither the number x nor the word "all"

were mentioned in the instructions. This rules out

the obvious: move all the chips to stack 3, then

move x of them to the middle.

## Recommended Posts

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He "handed a pile of poker chips to Jamie" to begin with. That means we're not talking about 1000 chips, maybe less than 100 chips.

So he can:
(one hundred times) Say "Move a chip from pile 2 to pile 1"
(one hundred times) Say "Move a chip from pile 3 to pile 1"
(X times) Say "Move a chip from pile 1 to pile 3"

If Ian can't obey an instruction because there are no more chips, Ian simply does nothing.

Somehow, I expect you want something smarter...

Edited by CaptainEd
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(one hundred times) Say "Move a chip from pile 2 to pile 1"

(one hundred times) Say "Move a chip from pile 3 to pile 1"
(X times) Say "Move a chip from pile 1 to pile 2"
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Jamie stacks A: 4+y B:4+y C:4+y
Davey did not call 1
Alex instruct :
take 1 from A ,shift to B : 4X
take 1 from C ,shift to B : 4X
Ian piles A: y B:12+y C:y
Alex instruct :
leave 1 on B ,shift to C
A: y B:1 C:11+2y
leave 1 on C ,shift to A
A: 10+3y B:1 C:1
take 1 from C ,shift to B
A: 10+3y B:2 C:0
leave 1 on A ,shift to C
A: 1 B:2 C:9+3y
take 1 from A ,shift to B
A: 0 B:3* C:9+3y

done if Davey called 3..but
continue to x if not
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Seems to me that TimeSpaceLighForce's would work, but moving all but 1 chip from the pile is basically the same as saying move all the chips. BTW, an easier and faster solution would be:

Move 4 chips from #1 to #3

Move all but 1 chip from #2 to #3

Move (12-x) chips from #3 to #2

But again too simpla a solution for a Bonavona puzzle

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Ooops, last move should be(x-1) chips to #2.

Another simple solution is for Monty to give Ian 12,13,or 14 chips, making 3 stacks of 4, leading to easy solution

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There are a number of trivial solutions that could fit the OP, like saying "move half of the chips in the middle pile into the leftmost stack, and move the other half into the rightmost stack" and then saying "move a chip from the left (or right) stack to the middle stack)" X times.

I think we might need more details about what sorts of instructions are legal in this problem, and what sorts aren't.

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I had in mind finding a method that did not seem to depend on

knowing the value of x. There may be very many such ways.

So let's amend the trick to make it seem more difficult,

but actually providing a guide to the desired solution.

1. Have Jamie construct three equal stacks.
You do not know how many chips are in each stack.

2. Instruct Ian to make three adjustments to the stacks.
The words all, half, etc may not be used; at no point is any stack empty.

3. Ask Davey to call out a number 0 < x < 13.
4. Instruct Ian to make a fourth adjustment to the stacks:
"Move __ chips from Stack _ to Stack _."
Where "__" is a specific number different from x.

The middle stack should now have x chips.

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Assuming that under the new rules the initial stacks must be at least 5 chips tall... (with only 12 chips in play it's impossible to guarantee that "at no point is any stack empty" and give a fourth instruction when x=12)

First 3 instructions:

1. Move 4 chips from stack 1 to stack 2

2. Move 4 chips from stack 3 to stack 2.

3. Move the number of chips in stack 1 minus 1 from stack 2 to stack 3

At this point the stack 2 will contain 13 chips.

Last instruction:

4. Move 13-x chips from stack 2 to stack 1.

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Several solution have been posted.

Mine is not greatly distinguished from any of them.

It differs from k-man's (above) only in that it works with 12 chips.

First three steps:

Take 3 chips from each end pile and add to the center.

Move from the center to one end the number of chip in either end pile.

There will now be nine chips in the center pile.

The last step increases or decreases the center pile as needed to get it to x.

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