<p>Assume that you have six coins, one of which is fake. The fake one weighs a little more or less than the others, but we don't know which. With only two weighings of scale, how can we identify the fake coin? (any change to the scale is considered a move)</p>
<p>Put two coins on each side. If one side of the scale goes down, take those two coins and re-weigh them to see which one is heavier. If they are the same, weigh the two remaining coins.</p>
<p>He said the fake coin could EITHER be heavier or lighter, we don't know which.</p>
<p>plus that's three weighings</p>
<p>piemastarr,
theblumuffin's solution consists of only 2 weighings.</p>
<p>And the EITHER heavier or lighter thing gives this problem a twist...I say get a bucket of water, throw the coins in, and see which coin has a different sinking rate compared to all others. Lol, a bit cheating, but that gets the stuff done in < 1 weighing.</p>
<p>Well the solution is obviously magic, then.</p>
<p>lalallalala</p>
<p>soooo we have six coins.. let's call them coin a, coin b, c, d, e, f =)</p>
<p>weight a+B with c+D;
if they're the same, thennnnnnnnnnnnnnn, we know that the fake one is either e or f and that a,b,c,d are allllllll reall
soo we just weigh one of a,b,c,d with e, if it's the same, then f is fake, if it's not the same, then e is fake</p>
<p>SO IF A+B and C+D ARE NOT THE SAME, this means E and F are both real, and one of A,B,C,D is FAKE.</p>
<p>we remeasure using</p>
<p>A+C on one side and B+D on another</p>
<p>and we're done.</p>
<p>how will this get u into harvard?</p>
<p>collegebond, didn't you get this test in your Harvard interview? I thought this was the standard for an automatic 'accept'.</p>
<p>foolonthehill161, your solution doesn't distinguish between a-heavy, d-light (or the opposite), or b-heavy, c-light (or the opposite). Try it out. I can get a solution that does a bit better, but it can't do all possibilities.</p>
<p>Indeed, I'm pretty sure that you can't actually do it in two weighings. In general, if you do n weighings, you can only distinguish between (3^n-3)/2 coins, which would be 3 in this case. The normal way this question is posed is for 12 coins with 3 weighings, which is a pretty simple problem.</p>
<p>ouch, very nice ;)</p>
<p>Actually, I misread the problem. I thought that we needed to figure out if the coin was heavier or lighter. It may be possible to identify which coin of six is fake with two weighings, though I don't see how to do it immediately.</p>
<p>i don't think it's possible with 2 scale weighings</p>
<p>
I don't think it's possible either. I was just trying to say that the maximum number of determinable coins (1/2(3^n-3)) is slightly off if you don't need to know the weight. For example, I know you can do 4 coins with 2 weighings and 13 coins with 3 weighings so it's possible that the maximum number of coins is (3^n-3)/2 + 1, but I don't have a proof for that.</p>
<p>any more takers?</p>
<p>A. You put two coins on each side.</p>
<p>If either side weighs more or less than the other, then you take out one coin from each side. </p>
<p>The result could be:
1) The scale balances out.
2) Either side still weighs more or less than the other.</p>
<p>If it's #1, then the coin that you took out from the side that weighed more or less is the fake one.</p>
<p>If it's #2, then the coin left on the scale that weighs more or less than the other is the fake one.</p>
<p>B. If step A results in a balanced scale, then you put the remaining two coins on each side.</p>
<p>The balance of the scale is bound to change, so the side that weighs more or less than the other after step B has the fake coin on top.</p>
<p>Is this it?</p>
<p>
[QUOTE]
A. You put two coins on each side.</p>
<p>If either side weighs more or less than the other, then you take out one coin from each side.</p>
<p>The result could be:
1) The scale balances out.
2) Either side still weighs more or less than the other.</p>
<p>If it's #1, then the coin that you took out from the side that weighed more or less is the fake one.</p>
<p>If it's #2, then the coin left on the scale that weighs more or less than the other is the fake one.</p>
<p>B. If step A results in a balanced scale, then you put the remaining two coins on each side.</p>
<p>The balance of the scale is bound to change, so the side that weighs more or less than the other after step B has the fake coin on top.</p>
<p>Is this it?
[/QUOTE]
</p>
<p>I don't know what the answer is, but that solution doesn't work, for a simple reason: you say " the coin left on the scale that weighs more or less than the other" and repeat this several times- but this is meaningless! If the scale isn't equal, then there isn't ONE side that weighs "more or less" than the other- one weighs more, one weighs less, and it's impossible to tell the difference in your model.</p>
<p>Think about step B- if you have the remaining two coins, then one will way more, one will weigh less. Which one is the coin that weighs more or less? It's impossible to tell. (By the way, you can replace step B with a step that works- measure ONE of the remaining coins with one of the coins you already weighed. If they're equal, then it's the other coin you didn't weigh, if they're unequal, it's the remaining coin that you did weigh).</p>
<p>Well, step B really isn't a problem, because to find the fake coin among the last two coins, he just has to compare one of those two coins with one of the 4 definitely real coins and see if they're the same. The real problem, as Admiral stated, is in this part:</p>
<p>"If it's #1, then the coin that you took out from the side that weighed more or less is the fake one.</p>
<p>If it's #2, then the coin left on the scale that weighs more or less than the other is the fake one."</p>
<p>We are always left with two solutions for both results. How do we know whether the fake coin was heavier or lighter? (ex: we won't know if we took a heavier fake coin from one side, or a lighter fake coin from the other side). This was actually what cghen was saying earlier.</p>
<p>Hm...</p>
<p>Does anyone think that this problem has a very simple answer?</p>
<p>I guess the original poster can surprise us...</p>
<p>We take the 6 coins and the scale to a coin expert. We offer to give him the scale if he tells us which coin is the fake in return.</p>