watchmath
  • watchmath
Suppose there are 100 coins that are arranged into 10 piles of coins each pile consist of 10 coins. There is exactly one pile where all the coins are fake and all coins on the other pile are genuine. Assuming that the weight of the fake coin is different from a genuine one, what is the minimal number of weighing to determine which pile is the fake one (say here you have a digital weighing machine ).
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I like this version more, because you don't know if it's heavier or lighter, which usually catches a few out/makes them miss the best method! I'll sit back and watch the lulz ensue.
anonymous
  • anonymous
if they're already in all-fake or all-genuine piles, then you essentially have 9 real and 1 fake. Weigh two sets of 3 against each other. If they're the same, it's in the other 4; then you weigh 1 from that pile against 1; if they're the same, weigh one against the third from the pile. If those are the same, it's the 4th one. If they're different, it's the 3rd. If when you weigh the first 2 of the 4 you get them to be different, weigh one against the third in the group; if these are the same then it was the other of the first two, and if they're different it was the one which was in both of the previous weighings Now, let's say the two sets of 3 have unequal weights. Weigh any 2 of those against another 2. If they're the same, it's in the OTHER 2, and so you weigh one of those four against one of the two; if it's the same, it's the other one of the 2, and if they're different, it's that one of the 2. Now, let's say you've weighed the threes and the twos to be unequal. Now you go through the same process as in the second paragraph with these four. And so, in the first 2 "scenarios" it takes 3 weighings, and the 3rd scenario takes 4. Perhaps there's a way to ALWAYS do it with 3? I don't know.
watchmath
  • watchmath
Hi, thebestpig that is the other type of coin problem. Here we have a digital weighing machine that tell you the weight of objects that you put on top of it.

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anonymous
  • anonymous
With *12 coins*, you can always do it with 3 *without* a digital weighing machine (though it's not easy to spot).
anonymous
  • anonymous
In fact, 'spot' is completely the wrong word.
anonymous
  • anonymous
Interestingly, the fact there are piles rather than just coins suggests there is a weird method involved; do you have a solution 'watchmath'?
watchmath
  • watchmath
of course :)
anonymous
  • anonymous
about the 12, is there a better first move than to weigh 4 against 4? Because if they're different, you've only narrowed it down to 8...
anonymous
  • anonymous
watchmath don't tell us yet I want to think a bit more
watchmath
  • watchmath
I won't ruin ruin the friday night of people who has no life here :D :D :D
anonymous
  • anonymous
(For the record, you're basing that assumption on me being in the American timezone, which isn't the case)
watchmath
  • watchmath
I am aware of that. I am not addressing all people here. I am just adressing people in US ( in particular rsvitale)
watchmath
  • watchmath
Please don't google because obviously someone already discuss this thousand of times.
anonymous
  • anonymous
no cheating would ruin the fun
watchmath
  • watchmath
I am sorry if I sound that I was adressing that to you. I see that several people look at this post, so I was adressing to all people.
anonymous
  • anonymous
I think you should actually say something like: 'I'd advise you not to google this, because it will take your fun out of it. However, I cannot stop you, so if you do google it do not come back her with the solutions, especially if you claim your did it yourself'
watchmath
  • watchmath
Thank you for saying that. My english is limited so I have problem to say a wordy statement like you said above :).
anonymous
  • anonymous
i think i can do it in 3. Is there a way to do it in less?
watchmath
  • watchmath
yes :)
anonymous
  • anonymous
I'm thinking about putting 1 from the first pile 2 from the second pile...10 from the 10th pile on the scale and thinking about what the total would tell you.
anonymous
  • anonymous
maybe that would only be helpful if we were dealing with integer weights though
anonymous
  • anonymous
well i give up for now, ill think about this later
watchmath
  • watchmath
ok, no need to rush.

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