minute by a factor of five. I will show that this reduction is still possible without knowing if the counterfeit coin is

lighter or heavier as long as only one possibility exists for each coin.

Consider five coins in which you have deduced that, of three, if one is the counterfeit, it must be lighter, and of the other

two, a counterfeit must be heavier. On each scale, you could take one coin from each group and weigh them together against

two known genuine coins, and set aside the fifth. A simpler solution is to weigh two light candidates against each other and

two heavy candidates against each other and set the fifth aside.

In two minutes, you can test 25 coins as long as you have some information on each coin. Put ten on each scale and set five

aside. The ten on each scale must be picked such that, if a scale becomes unbalanced, half of the coins will be proved

genuine.

In three minutes, 50 coins on each scale and set 25 coins aside for 125 coins, etc.

So when information is known before each weighing, the coins that can be tested in n minutes is 5^n.

If no information is known, then in one minute only 3 coins can be tested.

If you have two minutes and no information, put 5 coins on each scale and set 3 aside. 2*5+3=13 coins.

The five coins on a scale refers to five candidate coins. It may be three candidates in one pan and two candidates plus a

genuine coin on the other. Or it could be five candidate coins in one pan an five genuine in the other. If the scale tips,

information will be known about each candidate coin. Use the remaining weighings to reduce the set by five. If neither scale

tipped, the set of coins that was set aside must have the counterfeit and you have one minute less remaining and still no

information.

In three minutes, put 25 coins on each scale and set 13 aside. 2*25+13=63.

In four minutes, 2*125+63=313.

By now, maybe you see the recurrence f(n)=2*5^(n-1)+f(n-1), where f(1)=3, but not quite.

In five minutes, 2*624+313=1561 coins. <– final result.

(Not 2*625+313, since we have no genuine coins to offset an odd number of coins on a scale.)

So we start with 312 coins in each pan and 313 coins set aside. If a scale tips we know that 312 coins are light candidates

and 312 are heavy candidates. That can be reduced to 125, 25, 5, and 1 with the remaining four minutes.

If the scales remain balanced, divide the unknown coins as described and continue. ]]>

“What is the largest number of coins N for which it is possible to find the fake coin in five minutes?”

It is possible to find the fake coin in one weighing no matter how many coins there are just by getting lucky and picking it as one of two.

It should be worded to the effect of “… for which it can be guaranteed to find the fake coin in …” ]]>

Good one! You manage to beat the restriction I conjectured because, during the three final rounds, it is meaningful whether a scale falls left, falls right or is balanced, and not just whether it’s balanced or unbalanced. So there are five possible outcomes in these rounds.

]]>Then having determined if the fake coin is ligher of heavier, we can divide the group containing the fake coin in five equal parts, four of them are put on the scales. We can thus reduce the pile containing the fake coin by a factor of 5. So, this method yields N = 6*5^3 = 750.

]]>you just need to find.

]]>With the classic 12-coin, 3-weighing problem, it makes a difference. If you need to know whether the fake coin is heavy or light, 12 coins is the maximum. If you only need to find the fake, you can handle 13 coins.

]]>In this solution the information content of each round comes from a space of three possible answers: both scales are even, the first is uneven or the second is uneven. The answer “both scales are uneven” is impossible. Reducing by a factor 3 is optimal if the space of answers has cardinal 3.

Beating 3^5 is only possible if we manage to invent a configuration in which the outcome “both scales are uneven” is possible, otherwise we can’t draw more information. Is it possible?

]]>At the final step, I think we need to weight a possible fake coin against a known good coin, as weighing the fake coin against a possible fake will only tell us that one of them is a fake, not which one.

Additionally, we can mix in some good coins if needed so that all of the piles divide evenly in half.

Step 5: Weigh two possible fakes against two good coins. Do not weigh one possible fake. (3 coins). Leaves one possible fake coin.

Step 4: Weigh 1 possible fake and 1 good coin vs 2 possible fakes on each scale. Do not weigh 3 coins. (9 coins)

Step 3: Weigh 4 possible fakes and 1 good coin vs 5 possible fakes. Do not weight 9 possible fakes. (27 coins).

Steps 2 and 1 involve using 81 and 243 coins. The maximum number is 3^5 = 243.

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