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watchmath
 5 years ago
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 ).
watchmath
 5 years ago
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 ).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0if they're already in allfake or allgenuine 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
 5 years ago
Best ResponseYou've already chosen the best response.0Hi, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0With *12 coins*, you can always do it with 3 *without* a digital weighing machine (though it's not easy to spot).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In fact, 'spot' is completely the wrong word.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Interestingly, the fact there are piles rather than just coins suggests there is a weird method involved; do you have a solution 'watchmath'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0about 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
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath don't tell us yet I want to think a bit more

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I won't ruin ruin the friday night of people who has no life here :D :D :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(For the record, you're basing that assumption on me being in the American timezone, which isn't the case)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I am aware of that. I am not addressing all people here. I am just adressing people in US ( in particular rsvitale)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Please don't google because obviously someone already discuss this thousand of times.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no cheating would ruin the fun

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for saying that. My english is limited so I have problem to say a wordy statement like you said above :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think i can do it in 3. Is there a way to do it in less?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0maybe that would only be helpful if we were dealing with integer weights though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i give up for now, ill think about this later
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