Foiled Again

[plainglazed]

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.

[Prime]

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.

First weight 8 against 8 coins, leaving 11 coins on the side. The variation where a set of 8 coins weighs less than the other one is staightforward. The lighter group of 8 can not have more than 1 heavy coin. Split that set into 4 and 4, weigh them against one another, and the lighter set of 4 is standard. Should they weigh the same -- all 8 are standard.

The case where both sets of 8 weight the same is more difficult. There are 3 possibilites: each of the equal sets of 8 have 2 doubly wrapped coins, or they each have one, or both have zero coins. Correspondingly, the set of 11 on the side has zero, 2, or 4 heavy coins.
Lets name one set of 8 W1, the other set of 8 -- W2, and the set of 11 coins left on the side -- S11.
Then the second weighting should be as follows:
Leave W1 in the pan and put 4 more coins from W2 into the same pan. Lets call those 4 -- W2(4). Put one coin from W2 into the other pan, lets call it W2(1). Add the entire S11 into that pan. That leaves 3 coins from W2 on the side. Lets call them W2(3).
Thus after the second weighting of 12 coins against 12, the solution is as follows:
1. W1 + W2(4) > (weighs more than) W2(1) + S11, then the entire S11 is guarantied to be standard.
2. W1 + W2(4) = W2(1) + S11, then W2(1) + W2(3) have to be standard.
3. W1 + W2(4) < W2(1) + S11, then W2(4) must have standard coins only.

Here is the reasoning behind that solution:

1, If S11 contains standard coins only, its side of the balance should always be lighter. That is because in that case W1 contains 2 doubly wrapped coins and the balance side of S11 + W2(1) can at most have one. Also, this is the only case when that side could be lighter because in the other cases S11 contains either 2 or 4 heavy coins. Thus when S11 + W2(1) is lighter, S11 must be all standard coins.

2. If S11 has 2 doubly wrapped coins then
a). to balance the scale W1 + W2(4) must have the remaining 2 heavy coins, which leaves W2(1) + W2(3) standard.
b). when S11 + W2(1) side is heavier, knowing that W1 contains 1 heavy coin, leaves W2(4) standard.

3. Finally, when S11 contains all 4 doubly wrapped coins making its side of balance heavier, we can safely pick W2(4) as a set of standard coins.

I am curious as to how this problem was constructed.

[plasmid]

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

For the first weighing, put 12 coins on one pan, 12 coins on the other, and leave 3 off the balance.

If one pan is lighter, then that pan must hold no more than 1 silver foiled coin. In that case, take the coins from the lighter pan and put 6 of those coins on each pan of the balance. Since there is at most one silver foiled coin in that group of 12, you know that the lightest pan (or both pans if they're equal) must not have a silver foiled coin and you're done.

If the original weighing was balanced, then you know that the coin distribution is either

1 – one silver foiled coin on each pan, and two silver foiled coins are among the three coins not on the balance

2 – two silver foiled coins on each pan, and no silver foiled coins off the balance

For the second weighing, pick one of the pans and put 6 coins on each side of the balance. If the distribution was #1 from above, one of the pans must be heavier and you will know that none of the coins on the lighter pan are silver foiled. If the distribution was #2 from above and you happened to have put both silver foiled coins on the same pan, then you again can be assured that the lighter pan has no silver foiled coins (without knowing a priori whether you're dealing with distribution 1 or 2). If the pans are balanced, then you know that you must have been dealing with distribution 2 and that all three coins that were off the balance during the initial weighing do not have silver foil.

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.

[TimeSpaceLightForce]

If Tilted
13 < 13*** 1*
13* < 13** 1*
13* < 13*** 1
13 < 13**** 1
try lighter
6 < 6 * 1
6 = 6 1
6 = 6 1*

If Balanced
13** = 13** 1
try any
5* = 5 * 3
5 < 5* 3*

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

i like puzzle solutions , thanks

[plainglazed]

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

[TimeSpaceLightForce]

if tilted

8 < 8 11
try lighter

none or one
4 = 4
4 < 4'

if balanced

8 = 8 11
distribute by color
8' 4' = 1 11'' 3
8' 4 < 1' 11'' 3
8' 4 < 1 11'' 3'
8 4 < 1 11'''' 3
8'' 4'' > 1 11 3

Pls. note the previous post..TY

[TimeSpaceLightForce]

8'' 4 > 1 11 3"

missed again..

[TimeSpaceLightForce]

..more for completion

8'' 4' > 1' 11 3

8'' 4' > 1 11 3'

8' 4' > 1 11 3

8' 4 > 1 11 3'

From above broken posts :

If 8 4 ? 1 11 3

is balanced pick 1 + 3

if tilted right pick 4

if tilted left pick 4 /11

This confirms Prime's solution.. thanks

Edited December 17, 2012 by TimeSpaceLightForce

[Prime]

..more for completion

8'' 4' > 1' 11 3

8'' 4' > 1 11 3'

8' 4' > 1 11 3

8' 4 > 1 11 3'

From above broken posts :

If 8 4 ? 1 11 3

is balanced pick 1 + 3

if tilted right pick 4

if tilted left pick 4 /11

This confirms Prime's solution.. thanks

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

Complete variations for the 2nd weighing after the 1st weighing revealing two equal sets of 8:

8" + 4 > 1 + 11 aside 3"

8" + 4 > 1' + 11 aside 3'

8" + 4' > 1 + 11 aside 3'

8" + 4' > 1' + 11 aside 3

8" + 4" > 1 + 11 aside 3

Where set of 11 is always standard

8' + 4' = 1 + 11" aside 3

Where the sets of 1 and 3 are standard

8' + 4 < 1 + 11" aside 3'

8' + 4 < 1' + 11" aside 3

8 + 4 < 1 + 11"" aside 3

Where the set of 4 is always standard

[plainglazed]

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