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10-15-2010, 02:10 PM | #1 |
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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? |
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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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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10-15-2010, 02:57 PM | #4 | |
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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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10-15-2010, 03:13 PM | #5 | |
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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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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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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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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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10-30-2010, 07:15 AM | #9 | |
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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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11-02-2010, 03:17 PM | #10 |
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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. |
11-02-2010, 05:27 PM | #11 |
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I was a history major......
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01-23-2011, 07:38 PM | #12 |
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My idea:
#1-Divide the 100 coins into piles of 50. The heavier pile has the counterfeit. #2-Divide the heavier pile into piles of 25. The heavier pile has the counterfeit. #3-Divide the heavier pile into two piles of 12 (24). You hold one coin. IF the piles are equal, you have the counterfeit! IF they are unequal, you have a "normal" coin, and the heavier pile has the counterfeit. #4-Take the heavier pile (unless you were lucky above) of 12, and divide it into 6. The heavier pile has the counterfeit. #5-Take the heavier pile of 6 and divide into two groups of 3. The heavier pile has the counterfeit. #6-Take the heavier pile of 3, and put one coin on each pan. You hold one coin. IF the coins are equal-you have the counterfeit in your hand. IF the coins are unequal-that heavy coin is the counterfeit.
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03-03-2011, 11:54 PM | #13 |
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Another please. Since I missed out on this one. (Not that I would have come up with the right answer.)
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05-08-2011, 11:55 AM | #14 |
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You would separate the coins into four groups of 25. Put group #1 on one side of the scale and weigh group #2 against it.
1) First, we need to isolate a control group that has all good coins in it. Case A) If they weigh the same, neither has the bad coin. weigh group #3 against #1. If #1 and #3 are the same, then #4 is the culprit. If they are different, #3 is the culprit. Case B) If they weigh different, weigh group #3 against group #1. If they are different also, then #1 is the culprit. Take 5 coins from the good coins to be your control group and put the rest of the good coins aside to purchase ice cream and dark chocolate. 2) Whittle down some more. Break the culprit group into groups of 5. Put your control group on one side of the scale. Measure each of the culprit groups against the control group, stopping as soon as you find a group that is different. Save one coin from the control group and the culprit group that was different. Put the rest of the coins towards that new dildo you were saving for. 3) Go for the kill. Take one coin from the control group and measure it, one by one, against the coins from the new culprit group. As soon as you find a coin that does not weigh the same, you have found the counterfeit. Take the control coin and the other good coins and save up for some nice cologne for the boi who just found the counterfeit coin for you...and, please, remember to share the chocolate, ice cream, and dildo with the boi as well. |
05-08-2011, 12:00 PM | #15 | |
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10-20-2013, 09:09 AM | #16 |
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I'd just use my hands and see if there was any discernible difference before spending money on a scale. Then, I would compare it against another coin of the same denomination for a control.
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02-07-2015, 05:34 PM | #17 |
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I think people should remember the accuracy of their statements are so easy to check these days with all the modern technology available!
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02-07-2015, 06:31 PM | #18 |
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This is a variation of an interview question that I ask. I believe the optimal algorithm is (3) piles of 33 with one coin leftover. Do you have the actual answer?
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