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rodrigo
rodrigo
  • 121
  • 4

Explanation

The trick to solve this kind of puzzles is to think on the avaliable information, that is the possible outcomes of the weighings versus the possible solutions at each step.

A single weighing can go 3 different ways, left, right or balanced, so 5 weighings can go $3^5 = 243$ ways. The possible solutions of the problem are 121 heavier coins plus 121 lighter coins plus 1 no-fake equals 243. That means that with less than 5 weighing it is impossible to do, with 5 it might be possible.

Then, for each step, check that the possible solutions are never more than the possible outcomes. The possible solutions are on each step are the powers of 3: 243, 81, 27, 9 and 3.

For example, Weight #ED-3, there are 14 of one weight and 13 of the other. Those add up to 27 solutions, just right! We must separate 3 groups of 9 solutions each, so we leave 9 coins out, we change plates on 9 coins, and leave 9 coins unmoved. Then the next step will have 9 solutions in every case and all will go smoothly to the next step.

Explanation

The trick to solve this kind of puzzles is to think on the avaliable information, that is the possible outcomes of the weighings versus the possible solutions at each step.

A single weighing can go 3 different ways, left, right or balanced, so 5 weighings can go $3^5 = 243$ ways. The possible solutions of the problem are 121 heavier coins plus 121 lighter coins plus 1 no-fake equals 243. That means that with less than 5 weighing it is impossible to do, with 5 it might be possible.

Then, for each step, check that the possible solutions are never more than the possible outcomes. The possible solutions are on each step are the powers of 3: 243, 81, 27, 9 and 3.

For example, Weight #ED-3, there are 14 of one weight and 13 of the other. Those add up to 27 solutions, just right! We must separate 3 groups of 9 solutions each, so we leave 9 coins out, we change plates on 9 coins, and leave 9 coins unmoved. Then the next step will have 9 solutions in every case and all will go smoothly to the next step.

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rodrigo
rodrigo
  • 121
  • 4

I need 5 weighings, and I don't need the spare coin, thank you very much.

4041 coins on one plate, 40 coins on the other plus the known one. Leave 40 coins out

If it balances the fake is int the 40 left out, go to #E-2, if. If not go to #D-2.

4041 coins are of one weight, 4041 of the other.

Put 13 heavier plus 1314 lighter plus one known good on the left plate, 14 lighter plus 13 heavier on the right plate. Leave 14 heavier and 13 lighter out.

If it balances then the fake one is in the left out, 14 heavier and 13 lighter. If it goes to the left, the fake was unmoved, so it is one of the 13 heavier plus 14 lighter. If it goes to the right, thathe fake was moved, so it is one one of the 1314 lighter plus 13 heavier.

(The 13+13 may be consideded a subcase of 14+13).

I need 5 weighings, and I don't need the spare coin, thank you very much.

40 coins on one plate, 40 coins on the other.

If it balances go to #E-2, if not go to #D-2

40 coins are of one weight, 40 of the other.

Put 13 heavier plus 13 lighter plus one known good on the left plate, 14 lighter plus 13 heavier on the right plate. Leave 14 heavier and 13 lighter out.

If it balances then the fake one is in the left out, 14 heavier and 13 lighter. If it goes to the left, the fake was unmoved, so it is one of the 13 heavier plus 14 lighter. If it goes to the right, tha fake was moved, so it is one of the 13 lighter plus 13 heavier.

(The 13+13 may be consideded a subcase of 14+13).

I need 5 weighings.

41 coins on one plate, 40 coins on the other plus the known one. Leave 40 coins out

If it balances the fake is int the 40 left out, go to #E-2. If not go to #D-2.

41 coins are of one weight, 41 of the other.

Put 13 heavier plus 14 lighter on the left plate, 14 lighter plus 13 heavier on the right plate. Leave 14 heavier and 13 lighter out.

If it balances then the fake one is in the left out, 14 heavier and 13 lighter. If it goes to the left, the fake was unmoved, so it is one of the 13 heavier plus 14 lighter. If it goes to the right, the fake was moved, so it is one of the 14 lighter plus 13 heavier.

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rodrigo
rodrigo
  • 121
  • 4

I need 5 weighings, and I don't need the spare coin, thank you very much.

Where appropriate I'll assume that the scale will tip to one side and not the other, but because of simmetry it will be an equivalent reasoning for the other side. I will signal those places with (X)

Weighing #1

40 coins on one plate, 40 coins on the other.

If it balances go to #E-2, if not go to #D-2

Weighing #E-2

We know that the fake is in the remaining 40 coins or does not exist.

Of the 40 candidates, put 14 on one plate, 13 on the other plus 1 of the known good ones. Leave 13 out.

If it balances go to #EE-3, if not go to #ED-3

Weighing #EE-3

The fake one if present is in the 13 untouched coins.

We put 5 on one plate, 4 on the other plus 1 known good. Leave 4 out.

If it balances go to #EEE-4. If not there are 5 of one weight and 4 of the other. Go to #EED-4.

Weighing #EEE-4

The fake, if present is in the untouched 4 coins.

We put 2 on one plate, 1 on the other plus 1 known good. Leave 1 out.

If it balances go to #EEEE-5, if not go to #EEED-5.

Weighing #EEEE-5

The fake, if present is in the last untouched coin.

We put that single coin on one plate, a known good on the other.

If it balances no fake, if not, that one is the fake.

Weighing #EEED-5

Assume the scale will go to the left (X). Then the fake one is either one of the 2 left, heavier, or the other one, lighter.

We put 1 heavier coin on one plate, the other heavier coin on the other plate.

If it balances the fake is the lighter one. If not, the fake is the one that goes down.

Weighing #EED-4

5 coins are of one weight, 4 of the other. Assume that the 5 are heavier (X).

We put 2 heavier and 2 lighter on the left plate, 1 heavier, 1 ligther and 2 knwon ones on the other. Leave 2 heavier and 1 lighter out.

If it balances the fake is one of the 2 heavier and 1 lighter left out. If it goes to the right, the fake has changed plate, so is one of the 2 lighter plus 1 heavier that moved. If it goes to the left, the fake did not move, so it is one of the 2 heavier plust 1 lighter that did not move.

Go to #EEDD-5.

Weighing #EEDD-5

There are 2 candidates of one weight and 1 of the other. Assume that the 2 are heavier (X).

Put 1 of the heavier one plate and the other in the other plate.

If they balance, the fake one is the other one. If not, the fake is the one that goes down.

Weighing #ED-3

14 coins are of one weight, 13 of the other. Assume that the 14 are heavier (X).

Put 5 heavier plus 5 lighter on the left plate, 4 heavier plus 4 lighter plus 2 known good on the other. Leave 5 heavier and 4 lighter out.

If they balance the fake is in the left out, and there are 5 heavier and 4 lighter. Go to #EDD-4. If it goes to the left, the fake is in one that did not move, that is one of the 5 heavier plus 4 lighter. If it goes to the right, the fake is in one that moved, that is one of the 5 lighter plust 4 heavier.

Go to #EED-4

Weighing #D-2

40 coins are of one weight, 40 of the other.

Put 13 heavier plus 13 lighter plus one known good on the left plate, 14 lighter plus 13 heavier on the right plate. Leave 14 heavier and 13 lighter out.

If it balances then the fake one is in the left out, 14 heavier and 13 lighter. If it goes to the left, the fake was unmoved, so it is one of the 13 heavier plus 14 lighter. If it goes to the right, tha fake was moved, so it is one of the 13 lighter plus 13 heavier.

(The 13+13 may be consideded a subcase of 14+13).

Go to #ED-3

And that's it!!! I think that I covered all cases, if not, let me know and I'll review.