Thieving coin maker

colinwbs
Posers and Puzzles 17 Jun '12 20:44
1. 17 Jun '12 20:44
A medieval king had twenty gold coin makers.
He noticed that one was making coins one gram too light and pocketing the difference.
He also found a set of electronic scales which would allow him to take one measurement and then would self-destruct. (if 007 can, so can medieval kings)
All the royal coin makers lined up with a bag of their coins, the king took them, took one measurement and then executed one of his coin makers.

How did he know who to execute?
2. forkedknight
Defend the Universe
17 Jun '12 21:51
Originally posted by colinwbs
A medieval king had twenty gold coin makers.
He noticed that one was making coins one gram too light and pocketing the difference.
He also found a set of electronic scales which would allow him to take one measurement and then would self-destruct. (if 007 can, so can medieval kings)
All the royal coin makers lined up with a bag of their coins, the king t ...[text shortened]... ook one measurement and then executed one of his coin makers.

How did he know who to execute?
Number the coin makers 0 - 19

Take the nth coin maker and place n coins on the scale.

Take the expected weight - actual weight.

You will get the number of the coin maker responsible in grams.
3. sonhouse
Fast and Curious
21 Jun '12 10:00
Originally posted by forkedknight
Number the coin makers 0 - 19

Take the nth coin maker and place n coins on the scale.

Take the expected weight - actual weight.

You will get the number of the coin maker responsible in grams.
Me no understand the opening conditions here. Are you saying a random selection and then mixing of the coins together? It seems clear if you had a stack of coins, say 10 each, from each maker twenty measurements for sure would find the thief but knowing how much the coins are expected to weigh could find the thief on measurement #1 or measurement #20 by that method. So how are you selecting the coins to be measured?
4. 21 Jun '12 12:35
Originally posted by sonhouse
Me no understand the opening conditions here. Are you saying a random selection and then mixing of the coins together? It seems clear if you had a stack of coins, say 10 each, from each maker twenty measurements for sure would find the thief but knowing how much the coins are expected to weigh could find the thief on measurement #1 or measurement #20 by that method. So how are you selecting the coins to be measured?
Take 1 coin from minter #1.
Take 2 coins from minter #2.
Take 3 coins from minter #3.
Et cetera, until you get 20 coins from #20.

Put all these coins together. There are 210 coins in total. If all were honest, the pile would weigh 210 times the weight of one coin.

However, if minter 1 is fraudulent, 1 of the coins weighs 1 gram too little, and so does the whole pile.
If minter #2 is fraudulent, 2 coins weigh 1 gram too little, and the pile weighs 2 grams less than it should.
If minter #3 is fraudulent, 3 coins weigh 1 gram too little each, and the whole pile weighs 3 grams less than it should.
And so on, until if #20 is fraudulent, the pile comes up 20 grams too light.

Weigh the entire pile in one go. The amount of weight it comes up short, in grams, is the number of the minter who is cheating you.

Presuming, of course, that there isn't more than one minger among your minters.

(Or, of course, you could just get samples from them all and compare the weight by hand - one gram is a pretty big deficiency on a single coin. For some small ones, such as groats, it could be a quarter of the normal weight!)

Richard
5. sonhouse
Fast and Curious
21 Jun '12 17:01
Originally posted by Shallow Blue
Take 1 coin from minter #1.
Take 2 coins from minter #2.
Take 3 coins from minter #3.
Et cetera, until you get 20 coins from #20.

Put all these coins together. There are 210 coins in total. If all were honest, the pile would weigh 210 times the weight of one coin.

However, if minter 1 is fraudulent, 1 of the coins weighs 1 gram too little, and ...[text shortened]... n. For some small ones, such as groats, it could be a quarter of the normal weight!)

Richard
Bloody clever! Have to remember that!