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Riddle #11: Counterfeit Coin Revisited

Discussion in 'BBS Hangout' started by cmellon, Feb 20, 2003.

  1. cmellon

    cmellon Member

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    Warning: Don't read other posters' comments unless you give up since it usually contains the answer.

    Counterfeit Coin has come back to haunt you. :)

    Recall Riddle #5 question: a soon-to-be-executed mathematician is given one last chance if he is able to answer the riddle.

    He is presented with 9 coins, one of them is counterfeit. The counterfeit coin is lighter than the rest of the coins. He is given a balance scale and needs to find the counterfeit coin by using the scale "only twice".

    Invisible Fan has answered it:


    Now, this time the mathematician is presented with only 8 coins, one of them is counterfeit. However, this time the counterfeit coin can be lighter or heavier. Which one, the mathematician does not know.

    He is also given a balance scale. How many times using the balance scale can the mathematician find the counterfeit coin?
     
    #1 cmellon, Feb 20, 2003
  2. codell

    codell Member

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    Based on my theory, twice (based on actually loading it, not necessarily "using" it).

    Out of the 9 coins, take 8 of them and put them on the scale with 4 coins on either side. If the 2 sides are balanced then the coin that was not placed on the scale is the counterfeit.

    If the scale does not balance, keep removing a coin from each side until the scale becomes balanced. This would mean that one of the coins you removed is the counterfeit one. If the scale is still unbalances after removing all but one coin from each side, then the one of those two or the counterfeit.

    Once you have the two coins that are unbalances, weigh one of them against a known good coin. If the scale balances, its the coin that was not placed on the scale, if the scales dont balance, then its the other coin which was places on the scale.
     
    #2 codell, Feb 20, 2003
  3. cmellon

    cmellon Member

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    Actually as I stated, this time there are only 8 coins total.

    Each time you tinker the coin numbers on the scale, it's counted as "one time". So if you remove a coin from each side, it's the same as if you take the whole coins (8 coins total in your scenario), and put back 6 coins on the scale with 3 coins on either side.

    I'll give further help:
    Instead of asking you how many times can you find the counterfeit coin using the scale, I ask : How can you find the counterfeit one by using the balance scale "only three" times.
     
    #3 cmellon, Feb 20, 2003
  4. Grizzled

    Grizzled Member

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    Divide the coins up into groups of 2. Weight 2 of these groups against each other. If they do not balance then the counterfeit coin is one of those 4. If they do, it is one of the other four. Now balance two of the possible counterfeit coins against two that are known not to be counterfeit. If they balance, then the two possibilities are not counterfeit and one of the remaining two is. If the don’t balance then obviously one of the first two is the one. Now split up the pair known to contain the counterfeit. Weigh one against a coin that is known not to be counterfeit. If it doesn’t balance, it’s the one. If it does its pair is the one.
     
    #4 Grizzled, Feb 20, 2003
  5. cmellon

    cmellon Member

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    Yes, Grizzled got it. Move on to the next last riddle for today.
     
    #5 cmellon, Feb 20, 2003
  6. cmellon

    cmellon Member

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    Yes, Grizzled got it. Move on to the next last riddle for today.
     
    #6 cmellon, Feb 20, 2003
  7. Kimble14

    Kimble14 Member

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    I spent too much time on the presentation. D'oh! Now that I've typed it up, though, I'm not going to let it go to waste! Part 2 is basically Grizzled's answer; part 1 is the same, except broken out case by case so that you don't actually learn anything from the answer:

    (Part 1)
    Label the coins A-H.

    Weighing #1: AB vs. CD
    If #1 balances:
    . Weighing #2b: E vs. F
    . If #2b balances:
    . . Weighing #3bb: A vs. G
    . . If balances: H is counterfeit
    . . If not: G is counterfeit
    . If #2b doesn't balance:
    . . Weighing #3bu: A vs. E
    . . If balances: F is counterfeit
    . . If not: E is counterfeit

    If #1 doesn't balance:
    . Weighing #2u: A vs. B
    . If #2u balances:
    . . Weighing #3ub: E vs. C
    . . If balances: D is counterfeit
    . . If not: C is counterfeit
    . If #2u doesn't balance:
    . . Weighing #3uu: E vs. A
    . . If balances: B is counterfeit
    . . If not: A is counterfeit

    (Part 2)

    Step 1: Weigh 2 coins on each pan. If they balance, you'll know that they're all good. If not, you'll know that the other four are all good. Put the good ones in your pocket, because hey, you deserve it! Leave (or put) the others on the table the balance is presumably on; use the floor if necessary.

    Step 2: Weigh 1 coin from the table on each pan. If they balance, you know they're good, so pocket them. If not, pocket the other two on the table and put the ones on the scale back on the table.

    Step 3: Pull one of your good coins out and weight it against one of the 2 coins left on the table. If the scale balances, the coin on the table is counterfeit. If not, the one that's on the scale that you just got off the table is counterfeit.
     
    #7 Kimble14, Feb 20, 2003
  8. LeisureSuitMoochie

    LeisureSuitMoochie Member

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    Ok. I think I got it, but this will be really hard to explain...

    Start out with 4 groups of 2 coins apiece. Weight two of the groups together, if they weight the same (no counterfeit coins among the 4) then move on to the next 2 groups. If they are different weights, take two off and weigh them against each other. If these are different, weight one of them against an outside coin, if they are different, then that coin is the counterfeit, if they are the same, then the one you just took off is counterfeit. Use this same logic to figure out the other groups if two groups are ever the same weight.

    Confused?!? I thought so, I'll try and diagram it out...

    Number each coin 1-8.
    #1. Take coins 1 & 2 and weigh against coins 3 & 4. If they weigh differently, go to step #2. If they weigh the same, go to step #5.

    #2. Take coin 1 and weigh against coin 2. If they weigh differently, go to step #3. If they weigh the same, go to step #4.

    #3. Take coin 1 and weigh against any coin besides coin 2. If they weigh differently, coin 1 is the counterfeit. If they weigh the same, coin 2 is the counterfeit.

    #4. Take coin 3 and weigh against any coin besides coin 4. If they weigh differently, coin 3 is the counterfeit. If they weigh the same, coin 4 is the counterfeit.

    #5. Take coin 5 and weigh against coin 6. If they weigh differently, go to step #6. If they weigh the same, go to step #7.

    #6. Take coin 5 and weigh against any coin besides coin 6. If they weigh differently, coin 5 is the counterfeit. If they weigh the same, coin 6 is the counterfeit.

    #7. Take coin 7 and weigh against any coin besides coin 8. If they weigh differently, coin 7 is the counterfeit. If they weigh the same, coin 8 is the counterfeit.

    Whew! There you have it, any way (weigh) you measure, its only 3 times until you have found the counterfeit coin.
     
    #8 LeisureSuitMoochie, Feb 20, 2003
  9. cmellon

    cmellon Member

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    Hahaha... You get all the well-deserved credit for your nice presentation and answer Kim. ;)
     
    #9 cmellon, Feb 20, 2003
  10. cmellon

    cmellon Member

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    Another excellent explanation and solution by LeisureSuitMoochie. Thanks!
     
    #10 cmellon, Feb 20, 2003
  11. JuanValdez

    JuanValdez Member

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    LeisureSuitMoochie: great handle! :D
     
    #11 JuanValdez, Feb 21, 2003
  12. PhiSlammaJamma

    PhiSlammaJamma Member

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    Or he could have done this....


    Throw the balance against the wall bashing it into little parts. Take the parts and put them back together to make a whole. Then crawl out using the hole.
     
    #12 PhiSlammaJamma, Feb 21, 2003
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