The Leprechaun's Challenge

[Galendir]

While pulling up a piece of blackthorn root that you think would make a fine shillelagh, a glimmer of light from below the trunk catches your eye. As you investigate further, you find that there is a hidden hollow beneath the tree. Inside the hollow you find five variously sized chests each filled with gold coins. Suddenly a tiny red-bearded man dressed in green appears. “Ahh! You’ve come to steal my treasure!” He shouts. “You must think yourself pretty clever to have found my hidey-hole. Not so clever I’ll wager.” The little man snaps his fingers and suddenly a weighing scale appears at your feet. “Every genuine gold coin in this trove weighs precisely one ounce. But not all that glitters is gold, my friend,” he says with a smirk. “A false coin weighs exactly one hundredth of an ounce more or less than a true coin. All the coins in any single chest are identical. This scale will give you an accurate and precise reading. You may weigh whatever coins you like, but you may take only one reading from the scale. If you can correctly tell me the nature of coin in each chest, I’ll reward you with a bag of gold. Otherwise, I’ll put a curse on you.”


What strategy can you use to pass the leprechaun’s challenge and leave with a bag of gold?

[CliveStaples]

Just some musings....

You can't take one coin from each chest and then weigh a subset of them, because you won't know anything about the chests of the coins that you didn't weigh, and you won't know how any +/- 0.01 oz are distributed amongst the coins you did weigh. Any winning strategy will have to be agnostic WRT the number of authentic coins, since it's impossible to tell from one scale reading whether you had false coins whose errors cancelled or whether you have authentic coins.

[Ibelsd]

Just some musings....

You can't take one coin from each chest and then weigh a subset of them, because you won't know anything about the chests of the coins that you didn't weigh, and you won't know how any +/- 0.01 oz are distributed amongst the coins you did weigh. Any winning strategy will have to be agnostic WRT the number of authentic coins, since it's impossible to tell from one scale reading whether you had false coins whose errors cancelled or whether you have authentic coins.

Here is my take on this.

Actually, the puzzle says that each fake coin is +/- .01 of a real coin. However, it also says all the coins in an individual chest are identical. So, if one coin in a chest is -.01 of a real coin, then they must all be. Therefore, solving the puzzle should involve taking a set of coins from each chest such that if a chest holds fake coins will be identifiable based on the value away from the expected total it is. Chest 1: 1 coin Chest 2: 3 coins Chest 3: 5 coins Chest 4: 7 coins Chest 5: 9 coins Expected if all are real = 25 24.9 or 25.1 would mean all are real but chest 1. If the difference is exatly .3 then only chest two. If the difference is .4 or .2, then chest 1 and 2 are fake. And so on. I realize I am off by a factor of 10. I also believe the values may need to be prime numbers. So, instead of 9, then 11. I have not tested this all the way through.

[Squatch347]

I could sue him for plagiarizing Tolkien and then I wouldn't have to weigh the coins?

[Galendir]

Ibelsd, You are on the right track.

Squatch, I'm unfamiliar with anything remotely similar to this in any of Tolkien, but I've only read the Hobbit and part of the Silmarillion. (I've had the full set of Lord of the Rings since junior high and have never yet read it. Shameful, I know.) Could you provide a more specific reference to to the material you think this plagiarizes?

[Squatch347]

Squatch, I'm unfamiliar with anything remotely similar to this in any of Tolkien, but I've only read the Hobbit and part of the Silmarillion. (I've had the full set of Lord of the Rings since junior high and have never yet read it. Shameful, I know.) Could you provide a more specific reference to to the material you think this plagiarizes?

I was being somewhat facetious, sometimes the jokes fall a bit flat on the internet sorry. I meant to play off the phrase "But not all that glitters is gold..." which is similar to Tolkein's poem:

All that is gold does not glitter,
Not all those who wander are lost;
The old that is strong does not wither,
Deep roots are not reached by the frost.
From the ashes a fire shall be woken,
A light from the shadows shall spring;
Renewed shall be the blade that was broken,
The crownless again shall be king.

https://en.wikipedia.org/wiki/All_th...es_not_glitter

[Though to be appropriately critical of myself, I didn't realize the phrase originated with (or before) Aesop.]

[Galendir]

Solution

~ There are 3 possible types of coin per chest: True, Heavy-false, Light-false. For 5 chests that makes 3^5 or 243 total possible arrangements. Each arrangement can be expressed as a 5 digit ternary (trinary) number with each digit representing the type of coin in one of the chests. We can use the digits: L T H to represent Light, True or Heavy, or the symbols: -, 0, +. Number the chests 1 thru 5. Choose 1 coin from chest 1. The value of the first digit of our ternary number will represent the weight of this coin. If it weighs 0.99 oz (or -0.01 oz from true), it is light and represented by “-“, if it weighs 1 oz (or +/- 0 oz from true), it is true and represented by “0”, if it weighs 1.01 oz (or +0.01 oz from true), it is heavy and represented by “+”. Our second digit will represent the coin weight of chest 2. The value of this digit/column is three times that of the first digit/column so we use 3 coins. They will weigh: 2.97 oz (or -0.03 oz from true) = “-“, 3 oz (or +/- 0 oz from true) = “0”, or 3.03 oz (or +0.03 oz from true) = “+”. Our third digit will represent the coin weight of chest 3. The value of this digit/column is three times that of the second digit/column so we use 9 (3^2) coins. They will weigh: 8.91 oz (or -0.09 oz from true) = “-“, 9 oz (or +/- 0 oz from true) = “0”, or 9.09 oz (or +0.09 oz from true) = “+”. Our fourth digit will represent the coin weight of chest 4. The value of this digit/column is three times that of the third digit/column so we use 27 (3^3) coins. They will weigh: 26.73 oz (or -0.27 oz from true) = “-“, 27 oz (or +/- 0 oz from true) = “0”, or 27.27 oz (or +0.27 oz from true) = “+”. Our fifth digit will represent the coin weight of chest 5. The value of this digit/column is three times that of the fourth digit/column so we use 81 (3^4) coins. They will weigh: 80.19 oz (or -0.81 oz from true) = “-“, 81 oz (or +/- 0 oz from true) = “0”, or 81.81 oz (or +0.81 oz from true) = “+”. You weigh a total of 121 (1+3+9+27+81) coins generating a combined weight from 119.79 to 122.21 oz (or -1.21 to +1.21 oz from true) each with a unique solution. Let’s say the total weight of all 121 coins was exactly 120 oz. This is -100 hundredths of an ounce from true. That is, it is 100 hundredths of an ounce lighter than if all the coins were genuine gold coins. In our ternary notation (from right to left) this would be: --+0- or LLHTL if that’s easier to read. This represents the contents of the chests in reverse order: 54321. Chest 5 has light coins -0.81 oz, chest 4 has light coins -0.27 oz, chest 3 has heavy coins +0.09 oz, chest 2 has true coins +/- 0 oz, and chest 1 has light coins -0.01 oz. 121 oz -0.81 oz -0.27 oz +0.09 oz +0 oz -0.01 oz = 120 oz. __121.00 Expected weight if all were true coins _ -000.81 Chest 5 _ -000.27 Chest 4 _+000.09 Chest 3 __000.00 Chest 2 _ -000.01 Chest 1 __120.00 Total weight of coins weighed. This is the only solution that will give 120 oz. ~