There is a problem at Story's Chocolate Coins factory ...and at the busiest time of year ...and with their most popular product - the twenty seven piece gold coin purse. Apparently silver foiled coins are being misdirected to the gold foiling lines and being refoiled. Like the standard gold coins, these twice wrapped coins are perfectly identical to each other and visually identical to the standard gold coin in every way unless weighed; the second wrapping making them slightly heavier. Four such coins are some how finding their way into each twenty seven piece gold coin purse this season. Which just goes to show, not every Story's chocolate filled gold coin has a silver lining. All right, all right, the puzzle: using a balance scale twice, find four **standard** gold coins from one of this season's twenty seven piece gold coin purses.

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# Foiled Again

Best Answer Prime, 15 December 2012 - 06:54 PM

I have enjoyed this problem. It looks like something original in the old family of weighting problems. The solution I found seems to be the only solution.

I am curious as to how this problem was constructed.

Go to the full post### #1

Posted 09 December 2012 - 05:11 PM

### #2

Posted 14 December 2012 - 12:27 AM

I can come close, but not quite get this one yet. But to get started:

Are we allowed to use shenanigans like putting stacks of coins at different distances from the fulcrum of the balance? If so then it should become easier.

### #3

Posted 14 December 2012 - 07:18 PM

Hi ,i just joined this awsome place you have ..

i like puzzle solutions , thanks

### #4

Posted 14 December 2012 - 08:02 PM

hello plasmid - great start. CaptainEd offered a solution isolating one standard coin before we seemed to have lost a day of posts earlier in the week. in responding to him, had mentioned i thought this was a good puzzle. not mine but reinterpretted another involving more coins. thought maybe twenty seven would provoke maths substantiating no solution possible since most all of the coin weighing puzzles i've witnessed on this site tend to work out with powers of three. why i thought it was good was a seemingly natural progression from dumbfounded, to one coin, three, and now onto four. can assure you no shenanigans required. good luck.

and hello TimeSpaceLightForce - welcome to BrainDen! nice work in your above proposed solution. do think you omitted one possibility. if, after your first comparison balances, your second comparison could be 5* = 5 * 3 or 5 = 5 3**. hope you'll give it another go. cheers

### #5

Posted 15 December 2012 - 06:54 PM Best Answer

I have enjoyed this problem. It looks like something original in the old family of weighting problems. The solution I found seems to be the only solution.

I am curious as to how this problem was constructed.

Past prime, actually.

### #6

Posted 17 December 2012 - 07:39 AM

Pls. note the previous post..TY

### #7

Posted 17 December 2012 - 08:19 AM

missed again..

### #8

Posted 17 December 2012 - 05:53 PM

**Edited by TimeSpaceLightForce, 17 December 2012 - 05:54 PM.**

### #9

Posted 17 December 2012 - 09:09 PM

Spoiler for

Nice economical notation. For the sake of completness, the following is the enumeration for the 2nd weighing:

Past prime, actually.

### #10

Posted 18 December 2012 - 02:00 AM

Nice solve Prime and TimeSpaceLightForce. As i had mentioned, this is not an original puzzle of mine. It was "reworked" from an Alexander Chapovalov question on the 2010 Euler Math Olympiad where one standard coin in 100 is asked to be found and then the subsequent discussion on the max number of standard coins that can be found out of 100.

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