A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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 ).

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  3. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    With *12 coins*, you can always do it with 3 *without* a digital weighing machine (though it's not easy to spot).

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In fact, 'spot' is completely the wrong word.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Interestingly, the fact there are piles rather than just coins suggests there is a weird method involved; do you have a solution 'watchmath'?

  7. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    of course :)

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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...

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    watchmath don't tell us yet I want to think a bit more

  10. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I won't ruin ruin the friday night of people who has no life here :D :D :D

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (For the record, you're basing that assumption on me being in the American timezone, which isn't the case)

  12. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am aware of that. I am not addressing all people here. I am just adressing people in US ( in particular rsvitale)

  13. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Please don't google because obviously someone already discuss this thousand of times.

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no cheating would ruin the fun

  15. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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'

  17. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you for saying that. My english is limited so I have problem to say a wordy statement like you said above :).

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i think i can do it in 3. Is there a way to do it in less?

  19. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes :)

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe that would only be helpful if we were dealing with integer weights though

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well i give up for now, ill think about this later

  23. watchmath
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, no need to rush.

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.