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classic weighing

I know the question another way around: what is the minimal number of weighting runs that you need if you have n pills with one different?

common solution:
when having 3**n pills (or less), you need n runs.
case for n=1, 2:
for 3 pills:
one pill aside, one pill left, one pill right, either you see a difference in the balance or it is the pill aside.
for 9 pills:
first run: 3 pills aside, 3 pills left, 3 pills right, either you see a difference in the balance or it is among the 3 pills aside
second run: take the 3 concerning pills and proceed like n=1

So for 27 pills you need only 3 runs (9 pills aside, 9 pills left, 9 pills right, rest see above), for 81 pills only 4 runs etc. In my opinion more impressing than only 12 pills.



Markus Franke
Thursday, September 4, 2003

Solution : Classic Weighing

With all hopes to isolate the odd
Place (1,2,3 & 4) on left pan
And (5,6,7 & 8) on right pan
Keeping (9,10,11 & 12) out in a plate!
Try the FIRST weighing!

  In case the pans equaled
  Among (9,10,11 & 12) is the odd!
  Put (9,10 & 11) on the left pan
  and any three among 1 to 8 on the right pan.
  Try the SECOND weighing!
      Incase the pans equaled
      12 is the odd!
      Swallows the assassin
      and escapes in a hurry!
      In his imagination, the calm reader
      can compare the 12 with any other even
      and find it lighter or heavier in the FINAL weighing!
      In case the pans not equaled in the SECOND weighing
      Note(1) whether our pan(left) is  lower or higher
      to say the odd is heavier or lighter.
      Again put 9 on the left pan,10 on the right pan
      And keep the 11 out and try the FINAL weighing!
          If pans are equal 11 is the odd!
          If pans are not equal along with Note(1)
          Decide the odd either 9 or 10!

  In case the pans unequalled in the FIRST weighing
  Note(2) which one is lower
  Keeping (4,8,12) untouched
  Make a triangular change:
  (1,2,3) to the plate kept out
  (9,10,11) to the right pan
  and (5,6,7) to the left pan.
  and try the SECOND weighing!
      Now if the pans are  equal
      The odd is among (1,2,3)!
      If pans are unequal  but similar to Note(2)
      The odd is among (4,8)!
      If the pans are unequal but opposite to Note(2)
      The odd is among (5,6,7)!

      One more weighing the FINAL is at hand!
          Put to trial the three groups
          (1,2 & 3) , (4 & 8) and (5,6 & 7)
          as the group (9,10 & 11) was tried earlier!
          Note(2) of FIRST weighing and similar or opposite
          Notes of FINAL weighing assisting you,
          Have that single odd pill palmed!

SR Thiyagarajan

SR Thiyagarajan
Thursday, September 11, 2003

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