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Old 10-15-2010, 02:10 PM   #1
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Lightbulb What do you think?

My students obsessed about this problem today.

How would you solve it?


You have 100 valuable coins.

Well, one of the coins is counterfeit, so it’s not as valuable.

It’s either lighter or heavier than the genuine coins.

The only tool you have is a balance scale, and you have to pay a fee each time you use it.

How will you find the counterfeit coin?
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Old 10-15-2010, 02:24 PM   #2
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It depends if all the coins are of the same sise and ammout of value,some coins are made of diffent metals so weight is diffrent on each.
My soulution(sp?) would be to seprate all the coins to theire monatary amount value,then look at them very carefuly,then pick out the ones I think may not be real.Then drop them on the table one at the time till one dosent sound the same when it hits the table.Result is u dont need to use the scale at all to find the counterfit coin. u find the fake coin and save $$ not useing the scale.
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Old 10-15-2010, 02:32 PM   #3
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I'd invest $10 or so and go purchase another set of scales, one that didn't cost me money each time I used it.

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Old 10-15-2010, 02:57 PM   #4
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Quote:
Originally Posted by Rockinonahigh View Post
It depends if all the coins are of the same sise and ammout of value,some coins are made of diffent metals so weight is diffrent on each.
My soulution(sp?) would be to seprate all the coins to theire monatary amount value,then look at them very carefuly,then pick out the ones I think may not be real.Then drop them on the table one at the time till one dosent sound the same when it hits the table.Result is u dont need to use the scale at all to find the counterfit coin. u find the fake coin and save $$ not useing the scale.
The coins appear identical.

The only difference between the valuable real coins and the counterfeit coin is that

The fake coin is either heavier or lighter than the real coins.
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Old 10-15-2010, 03:13 PM   #5
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Originally Posted by Chancie View Post
The coins appear identical.

The only difference between the valuable real coins and the counterfeit coin is that

The fake coin is either heavier or lighter than the real coins.
Okay my process of elimination and I am sure there is a better way but,
I would put 50 coins on each side of the scale, then take either the lighter or heavier side and divide them into 25 coins, if they weighed even then the counterfeit coin is amongst the other 25 coins, if they weighed uneven the counterfeit coin is amongst that 25. I would continue dividing the coins into equal amounts until I was left with only the counterfeit coin. Or you could just bite them and see if you were left with teeth marks And if this makes no sense then I blame it on being aussie ty
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Old 10-15-2010, 03:13 PM   #6
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The difference between the real coins and the fake coin is too subtle to tell with your hands.

Your task is find a way to use the balance scale efficiently.

ETA: Dreamer and I posted at the same time.
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Old 10-15-2010, 03:56 PM   #7
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I am horrible at these kind of problems but I love them.Please post the answer even if no one gets it!
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Old 10-15-2010, 04:06 PM   #8
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This might as well be the "one train leaves the NY station at 10 am doing 100 mph, one train leaves Chicago at 10 am doing 150 mph, what time do they intersect in.....Timbucktoo?"
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Old 10-15-2010, 04:44 PM   #9
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Ooooo.....I love love LOVE these kinds of problems!!!!

Wish I'd gotten in on this one a bit sooner.

~Theo~ .....loves puzzles and problem-solving exercises.
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Old 10-15-2010, 05:19 PM   #10
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i imagine that halving them til you get to the winner is the better idea... if the scale has to be used and they are identical i believe that would be the best bet...
of course... i am sure that there must be something else that works better... because that just sounds too logical and easy...
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Old 10-15-2010, 05:28 PM   #11
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Take 1 known genuine coin by weight. Divide the rest into 3 of 33 and weight them. The pile that is lighter or heavier gets cut into 3rds till you single it out.
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Old 10-15-2010, 05:39 PM   #12
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Well the question is ambiguous .If you have to pay each time you use the scale, does putting them on and then off count as one time or two? Also if you have to pay everytime you use the scale what's the goal here, what's the object of this quest? To spend the least amount of money possible in your quest, or is the goal to find out which coin is bogus? Anyway I have about 20 more questions because your details are not very clear. But here's my answer. Assumming putting all the coins on the scale counts as a one time use , and taking them all off counts as an additional use, I'd put all the coins on the scale. ( that's one time) Then I'd take them off one at a time, watching the scale for the difference in weights. (For example the first 15 I took off, all weighted 3 ounces each but the 16th one weighted 2 ounces, and the next 4 I took off weighted 3 ounces, I'd set the one that weightes 2 ounces aside) I'd set the different one aside and finished taking the other coins off, that would count as the second use of the scale. So 2 times I'd use the scale. If putting them on and off counts as one time use then I only used the scale once.
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Old 10-15-2010, 05:59 PM   #13
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Math is hard.

I'd bring them all to a coin dealer and let him or her sort it out.

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Old 10-15-2010, 06:24 PM   #14
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nooooo... you have to use that scale!

Quote:
Originally Posted by Gemme View Post
Math is hard.

I'd bring them all to a coin dealer and let him or her sort it out.

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Old 10-15-2010, 06:29 PM   #15
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Quote:
Originally Posted by softness View Post
nooooo... you have to use that scale!
Make me.



*grin*

This is not my forte but I am interested in the best answer.
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Old 10-15-2010, 07:13 PM   #16
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Well the question is ambiguous .If you have to pay each time you use the scale, does putting them on and then off count as one time or two? Also if you have to pay everytime you use the scale what's the goal here, what's the object of this quest? To spend the least amount of money possible in your quest, or is the goal to find out which coin is bogus? Anyway I have about 20 more questions because your details are not very clear. But here's my answer. Assumming putting all the coins on the scale counts as a one time use , and taking them all off counts as an additional use, I'd put all the coins on the scale. ( that's one time) Then I'd take them off one at a time, watching the scale for the difference in weights. (For example the first 15 I took off, all weighted 3 ounces each but the 16th one weighted 2 ounces, and the next 4 I took off weighted 3 ounces, I'd set the one that weightes 2 ounces aside) I'd set the different one aside and finished taking the other coins off, that would count as the second use of the scale. So 2 times I'd use the scale. If putting them on and off counts as one time use then I only used the scale once.
It's a balance scale, so you can compare coins in groups or singly.

And the idea is to be efficient.

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Old 10-17-2010, 11:08 AM   #17
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There are a few different ways to start this problem.

My students agreed that

They would divide the coins in four piles with 25 coins each, and

I labeled them A, B, C, and D.

First we compared A and B, and either

They balanced the scale, which meant that they were all valuable coins or

They were different, which meant that all the coins in C and D were genuine.
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Old 10-18-2010, 09:40 PM   #18
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Yay for Dreamer being the closest.
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Old 10-30-2010, 07:15 AM   #19
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Quote:
Originally Posted by Chancie View Post
My students obsessed about this problem today.

How would you solve it?


You have 100 valuable coins.

Well, one of the coins is counterfeit, so itís not as valuable.

Itís either lighter or heavier than the genuine coins.

The only tool you have is a balance scale, and you have to pay a fee each time you use it.

How will you find the counterfeit coin?
Okay, I'm probably going to show my ass here but I'm going to respond before I read anyone's responses:

I would take the coins, divide them into four equal groups then compare G-1, G-2, G-3 and G-4.

First iteration: compare G-1 and G-2. If equal then go to second iteration.
Second iteration: compare G-3 and G-4 at this point ONE of the piles will be unequal because the conterfeit coin has to be in one of those groups.

That's the most elegant solution I can come up with--at least while I wait for the coffee to finish brewing.
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Old 11-02-2010, 03:17 PM   #20
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That is very like what my kids came up with as a group.

We had already talked about something our old textbook called the guessing game algorithm, and

That algorithm is a lot like binary search.

So they hypothesized that it would be efficient to divide the coins into groups, and

They knew that it wouldn't be useful to divide the coins into two groups because

The still wouldn't know which pile contained the counterfeit and

They wanted something like a control, and

Having the control changed the way they searched after the first few iterations.

Anyway, it's worth noting that though the honors class came up with the naming system quickly and

They were able to predict some of the sticky places,

The other ninth grade class did very well at working together to solve this problem.
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