<p>You have 12 coins, each looking exactly alike but one being infinitesimally lighter or heavier than the others. You have a scale balance and three weighings of the coins to determine which one is different, and whether it's heavier or lighter.</p>
<p>Apparently, this puzzle has puzzled people since ancient times.</p>
<p>I can only narrow it down to 2.</p>
<p>Ah I've seen the solution. It required a little trick I didn't see. :)</p>
<p>If you know whether or not the coin is lighter/heavier, this will work (I'm going to assume it's lighter):</p>
<p>Put 5 on each side. If they are equal, weigh the two that are off the scale, and take the lighter one. Else, take the 5 remaining and weigh 2 on each side. If they are equal, it's the 1 remaining, else weigh the lighter two. The lightest of the two is the winner then.</p>
<p>If you know whether the coin is lighter or heavier, it's trivial. My slightly more straightforward solution is -</p>
<p>Divide in two groups of 6 each and weigh em, thus narrowing to 6 coins. Then divide into 3-3 and weigh, narrowing to last 3. Now weigh any two coins of the 3. If they are the same weight, the remaining coin is the odd one. If not, the one which weighed lighter is the odd one...</p>
<p>But ofcourse the above is wrong :p
Coin can be lighter OR heavier (unknown), right?</p>
<p>Right. Otherwise, as you said, it's trivial: just a problem, not a puzzle. I did the puzzle, with much sweat, about a decade ago and no longer have the solution. The trick, as I recall, was to number the coins, divide them into three groups (of four), and then, of course, to weigh two of the groups in the balance. If they're equal, thus commences the 'easy' part of the problem. If they're not, you have your work cut out for you. At least I did, but then I'm not that smart or logical, and this is about pure logic and the rigorous exploitation of derived information.</p>
<p>dwharris has the right idea. I also did this problem a long time ago, but it took me a while to remember the solution. I hope this explanation works. (It would be better if I displayed it in a tree structure I guess, but Ill try paragraph form.)</p>
<p>Number the coins 1-12. Divide coins into 3 groups: 1-4 are group A, 5-8 are B, and 9-12 are C. The first weighing is group A vs. group B.</p>
<p>The easy case is when A=B, and you know the coin is in group C. For the 2nd weighing, weigh 2 coins in group C against each other, for example 9 vs. 10. If those are equal, you know it is either 11 or 12 that is the different one. For the last weighing, do 9 vs. 11. If equal, #12 is the answer. If not equal, its #11. If however, 9 vs. 10 was NOT equal, say 9 < 10, then it is either #9 and its lighter or #10 and its heavier. So for the last weighing do 9 vs. 11. If equal, then the answer is #10. If 9 < 11 then its #9 (Note that its not possible for you to get 9 > 11).</p>
<p>So that branch covered the solution being in group C. Now back to the original weighing of A vs. B, suppose its not equal and without loss of generality lets say A < B. Now we know the coin is either in group A and light or group B and heavy. For the 2nd weighing, put coins 1, 2, and 5 on the left and 4, 6, and 9 on the right (other configurations with this pattern work as well, but Im going to use those numbers).</p>
<p>If (1, 2, 5) = (4, 6, 9), then those coins are all ok, leaving the possibilities that 3 is light or that either 7 or 8 is heavy. So for the last weighing, do 7 vs. 8. If equal, #3 is the light coin. If 7 < 8, #8 is heavy. If 7 > 8, then #7 is heavy. </p>
<p>If (1, 2, 5) < (4, 6, 9), then the possibilities are that either 1 or 2 is light, or that 6 is heavy. So for the last weighing, do 1 vs. 2. If equal, #6 is the heavy coin. If not equal, whichever of 1 or 2 is lighter is the answer.</p>
<p>If (1, 2, 5) > (4, 6, 9), then the possibilities are that #5 is heavy, or that #4 is light. For the last weighing, do 1 vs. 4. If equal, #5 is the heavy coin. If 1 > 4, then #4 is light. (Note that its not possible for 1 < 4).</p>
<p>Ok, I have way too much time on my hands. :-P Did that make any sense?</p>
<p>If the coin for which you are looking is truly "infinitesimally" lighter or heavier, then numbering the coins could skew the results. The ink on coin 12 would weigh more than the ink on coin 1.</p>
<p><em>ducks</em></p>
<p>:-)</p>
<p>Use 12 colors of marker and put the same mark on each, just in 12 different colors.</p>
<p>mit_18c - In the easy case when A = B and for your first example where you end up finding that 12 is the different one after all the previous weighings were equal...you still didn't figure out if 12 was lighter or heavier</p>
<p>But that's not part of the problem, BigReDoG!</p>
<p>ben & mootmom, you two are awesome :-)</p>
<p>so the original post does say:</p>
<p>"determine which one is different, <em>and</em> whether it's heavier or lighter." [emphasis added by me]</p>
<p>so unless the problem statement is incorrect, BiGReDoG makes a valid point...this problem is harder than i thought!</p>
<p>Ok, so in light of what BiGReDoG pointed out, here is a corrected solution for when A=B in the first weighing.</p>
<p>For the 2nd weighing, put two coins from group C on one side, and on the other side put one coin from C and one coin not from C. So, for example, (9, 10) vs. (11, 5).</p>
<p>If (9, 10) = (11, 5), then #12 is the different coin. Weigh it against any other to find if it's lighter/heavier.</p>
<p>If (9, 10) < (11, 5), then either 9 or 10 is light, or 11 is heavy. For the last weighing do 9 vs. 10. If equal, #11 is heavy. If not equal, whichever of 9 or 10 is lighter is the answer.</p>
<p>If (9, 10) > (11, 5), then either 9 or 10 is heavy, or 11 is light. Almost exactly like the previous case, so do the same thing and weigh 9 vs. 10. If equal, #11 is light, if not equal then it's whichever one is heavier.</p>
<p>Phew !</p>
<p>Yes, I remember that solution MIT_18C. I remember that choosing the coins for the second weighing in the 'unequal' case was a bear. I did it by pure guessing. Did you use any principle to come up the coins? Or did you guess too?</p>