[Guest]

So it looks like there are several variations of this but none I have found meet these exact criteria. Its for a game I play where I'm having to solve several different logic problems and this one has got me stuck. Just to get to this there were 5 other variations starting with 9 coins all the way up to 13 with a real coin to test. Those I finally got through. Here is what I'm working on now...

For this challenge, you are presented with a set of 13 suspect coins. Of these, it

is known that the number of counterfeit coins is either 0 or 2. All of

the counterfeit coins, if any, are known to have come from a single batch

of heavy coins. You must identify the counterfeit coins, if any, after 4

weighings or less.

Now I've got it down to where I know I am making a very educated guess but the game is rigged to where if I don't weigh the coins to where there can be no doubt it will decided to change the answer so no getting lucky here.

Any ideas?

[Guest]

Actually here's some more...

When you were one-fifth

as old as you will be seven years from now, you were as old as you were

nine years before you were one-third as old as you will be eight years

from now.

For this

challenge, you have flask A, which holds 12 ounces, and flask B, which

holds 19 ounces. Using only the operations of filling a flask, emptying a

flask, and pouring one flask into another (until either the first is

empty or the second is full), you must manage to put exactly 9 ounces of

water in flask A, leaving flask B empty.

[Guest]

When you were one-fifth

as old as you will be seven years from now, you were as old as you were

nine years before you were one-third as old as you will be eight years

from now.

The equation is

(x+7)/5 = (x+8)/3-9

x+7 = 5(x+8)/3-45

x+52 = 5(x+8)/3

3x+156 = 5x + 40

2x = 116

x = 58

So you are now 58.

checking

(58+7)/5 = (58+8)/3-9

65/5 = 66/3-9

13 = 22-9 = 13

[Guest]

I refer to flask a as 12 and flask b as 19 below to make it easier to track which is which.

Fill flask with 12

Pour into 19. Fill 12 again, pour into 19. Have 5 left in 12

Fill 19, pour into 12, have 14 left in 19 jar

Empty 12, pour from 19 to 12, have 2 left in the 19 jar

Fill 19, pour into 12, have 9 in the 19 jar.

Empty the 12 jar, pour the 9 from 19 into it. Done

[Guest]

When you were one-fifth

as old as you will be seven years from now, you were as old as you were

nine years before you were one-third as old as you will be eight years

from now.

The equation is

(x+7)/5 = (x+8)/3-9

x+7 = 5(x+8)/3-45

x+52 = 5(x+8)/3

3x+156 = 5x + 40

2x = 116

x = 58

So you are now 58.

checking

(58+7)/5 = (58+8)/3-9

65/5 = 66/3-9

13 = 22-9 = 13

Perfect, thanks I suck at math.

[Guest]

I refer to flask a as 12 and flask b as 19 below to make it easier to track which is which.

Fill flask with 12

Pour into 19. Fill 12 again, pour into 19. Have 5 left in 12

Fill 19, pour into 12, have 14 left in 19 jar

Empty 12, pour from 19 to 12, have 2 left in the 19 jar

Fill 19, pour into 12, have 9 in the 19 jar.

Empty the 12 jar, pour the 9 from 19 into it. Done

Well it seems the game is a bit bugged. Once I got 14oz in b it would pour evenly into a. Which is not possible! Anyway I was able to get 2oz in b by another route then follow the rest of your steps and it worked perfectly. I've been racking my brain with these things for days and now it seems now obvious. I suppose I just don't have the quick problem solving skills of some.

Again thank you.

[plainglazed]

Compare 1,2,3,4 vs 5,6,7,8

If 1,2,3,4 > 5,6,7,8 - there must be two counterfeit coins. either one in 1,2,3,4 and the other in 9,10,11,12,13 or both in 1,2,3,4 with 5,6,7,8 normal (X)
Compare 1,2,9 vs 3,4,10

If 1,2,9 > 3,4,10 the possible pairs are 1,2 - 1,9 - 2,9 - 1,11 - 1,12 - 1,13 - 2,11 - 2,12 - 2,13
Compare 1,9,X vs 11,12,13

If 1,9,X > 11,12,13 the possible pairs are 1,9 - 1,2 - 9,2
Compare 1 vs 9. If 1>9 then 1,2 are the counterfeit coins. If 1<9 then 9,2 are the counterfeit coins. If 1=9 then 1,9 are the counterfeit coins.
If 1,9,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13
Compare 11 vs 12. If 11>12 then 2,11 are the counterfeit coins. If 11<12 then 2,12 are the counterfeit coins. If 11=12 then 2,13 are the counterfeit coins.
If 1,9,X = 11,12,13 the possible pairs are 1,11 - 1,12 - 1,13
Compare 11 vs 12. If 11>12 then 1,11 are the counterfeit coins. If 11<12 then 1,12 are the counterfeit coins. If 11=12 then 1,13 are the counterfeit coins.


If 1,2,9 < 3,4,10 the possible pairs are 3,4 - 3,10 - 4,10 - 3,11 - 3,12 - 3,13 - 4,11 - 4,12 - 4,13
Compare 3,10,X vs 11,12,13

If 3,10,X > 11,12,13 the possible pairs are 2,3 - 2,10 - 3,10
Compare 2 vs 3. If 2 > 3 then 2,10 are the counterfeit coins. If 2 < 3 then 3,10 are the counterfeit coins. If 2 = 3 then 2,3 are the counterfeit coins.
If 3,10,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13
Compare 11 vs 12. If 11 > 12 then 2,11 are the counterfeit coins. If 11 < 12 then 2,12 are the counterfeit coins. If 11 = 12 then 2,13 are the counterfeit coins.
If 3,10,X = 11,12,13 the possible pairs are 3,11 - 3,12 - 3,13
Compare 11 vs 12. If 11 > 12 then 3,11 are the counterfeit coins. If 11 < 12 then 3,12 are the counterfeit coins. If 11 = 12 then 3,13 are the counterfeit coins.


If 1,2,9 = 3,4,10 the possible pairs are 1,3 - 1,4 - 2,3 - 2,4 - 3,9 - 4,9 - 1,10 - 2,10
Compare 1,3,9 vs 2,4,X

If 1,3,9 > 2,4,x the possible pairs are 1,3 - 1,10 - 3,9
Compare 9 vs 10. If 9 > 10 then 3,9 are the counterfeit coins. If 9 < 10 then 1,10 are the counterfeit coins. If 9 = 10 then 1,3 are the counterfeit coins.
If 1,3,9 < 2,4,X then possible pairs are 2,4 or 2,10
Compare 4 vs 10. If 4 > 10 then 2,4 are the counterfeit coins. If 4 < 10 then 2,10 are the counterfeit coins.
If 1,3,9 = 2,4,X the possible pairs are 1,4 - 2,3 - 4,9
Compare 1 vs 9. If 1 > 9 then 1,4 are the counterfeit coins. If 1 < 9 then 4,9 are the counterfeit coins. If 1 = 9 then 2,3 are the counterfeit coins.

A similar analysis can be made if 1,2,3,4 < 5,6,7,8

If 1,2,3,4 = 5,6,7,8 - either one counterfeit coin is in 1,2,3,4 and the other in 5,6,7,8; both are in 9,10,11,12,13; or all are equal
Compare 1,2,5,9,10 vs 3,4,6,11,12

If 1,2,5,9,10 > 3,4,6,11,12 the possible pairs are 1,5 - 1,7 - 1,8 - 2,5 - 2,7 - 2,8 - 9,13 - 10,13
Compare 1,9,10 vs 5,7,8

If 1,9,10 > 5,7,8 the possible pairs are 10,13 or 9,13
Compare 10 vs 9. If 10 > 9 then 10,13 are the counterfeit coins. If 10 < 9 then 9,13 are the counterfeit coins.
If 1,9,10 < 5,7,8 the possible pairs are 2,5 - 2,7 - 2,8
Compare 5 vs 7. If 5 > 7 then 2,5 are the counterfeit coins. If 5 < 7 then 2,7 are the counterfeit coins. If 5 = 7 then 2,8 are the counterfeit coins.
If 1,9,10 = 5,7,8 the possible pairs are 1,5 - 1,7 - 1,8
Compare 5 vs 7. If 5 > 7 then 1,5 are the counterfeit coins. If 5 < 7 then 1,7 are the counterfeit coins. If 5 = 7 then 1,8 are the counterfeit coins.

If 1,2,5,9,10 < 3,4,6,11,12 the possible pairs are 1,5 - 1,7 - 1,8 - 2,5 - 2,7 - 2,8 - 9,13 - 10,13
Compare 3,11,12 vs 6,7,8

If 3,11,12 > 6,7,8 the possible pairs are 11,13 or 12,13
Compare 11 vs 12. If 11 > 12 then 11,13 are the counterfeit coins. If 11 < 12 then 12,13 are the counterfeit coins.
If 3,11,12 < 6,7,8 the possible pairs are 4,6 - 4,7 - 4,8
Compare 6 vs 7. If 6 > 7 then 4,6 are the counterfeit coins. If 6 < 7 then 4,7 are the counterfeit coins. If 6 = 7 then 4,8 are the counterfeit coins.
If 3,11,12 = 6,7,8 the possible pairs are 3,6 - 3,7 - 3,8
Compare 6 vs 7. If 6 > 7 then 3,6 are the counterfeit coins. If 6 < 7 then 3,7 are the counterfeit coins. If 6 = 7 then 3,8 are the counterfeit coins.

If 1,2,5,9,10 = 3,4,6,11,12 the possible pairs are 9,11 - 9,12 - 10,11 - 10,12 - 3,5 - 4,5 - 1,6 - 2,6 - or all are equal
Compare 1,3,9,11 vs 2,4,10,12

If 1,3,9,11 > 2,4,10,12 the possible pairs are 9,11 - 3,5 - 1,6
Compare 1 vs 3. If 1 > 3 then 1,6 are the counterfeit coins. If 1 < 3 then 3,5 are the counterfeit coins. If 1 = 3 then 9,11 are the counterfeit coins.
If 1,3,9,11 < 2,4,10,12 the possible pairs are 10,12 - 4,5 - 2,6
Compare 2 vs 4. If 2 > 4 then 2,6 are the counterfeit coins. If 2 < 4 then 4,5 are the counterfeit coins. If 2 = 4 then 10,12 are the counterfeit coins.
If 1,3,9,11 = 2,4,10,12 the possible pairs are 9,12 or 10,11 or all coins are equal.
Compare 9 vs 10. If 9 > 10 then 9,12 are the counterfeit coins. If 9 < 10 then 10,11 are the counterfeit coins. If 9 = 10 all coins weigh the same.

Pretty messy to type out and am sure to have made some typos if not poor logic all together. Do think in theory this puzzle is solvable. With four questions one can
discern 3^4 or 81 pieces of information using a balance scale (>,<,= with each balancing) and the total number of possibilities I believe would be 13C2 + 13C0 or 79.

Label coins 1-13.

[Guest]

Label coins 1-13.


Compare 1,2,3,4 vs 5,6,7,8


If 1,2,3,4 > 5,6,7,8 - there must be two counterfeit coins.  either one in 1,2,3,4 and the other in 9,10,11,12,13 or both in 1,2,3,4 with 5,6,7,8 normal (X)

 Compare 1,2,9 vs 3,4,10


  If 1,2,9 > 3,4,10 the possible pairs are 1,2 - 1,9 - 2,9 - 1,11 - 1,12 - 1,13 - 2,11 - 2,12 - 2,13

   Compare 1,9,X vs 11,12,13


    If 1,9,X > 11,12,13 the possible pairs are 1,9 - 1,2 - 9,2

     Compare 1 vs 9. If 1>9 then 1,2 are the counterfeit coins. If 1<9 then 9,2 are the counterfeit coins. If 1=9 then 1,9 are the counterfeit coins.

    If 1,9,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13

     Compare 11 vs 12.  If 11>12 then 2,11 are the counterfeit coins.  If 11<12 then 2,12 are the counterfeit coins.  If 11=12 then 2,13 are the counterfeit coins.

    If 1,9,X = 11,12,13 the possible pairs are 1,11 - 1,12 - 1,13

     Compare 11 vs 12.  If 11>12 then 1,11 are the counterfeit coins.  If 11<12 then 1,12 are the counterfeit coins.  If 11=12 then 1,13 are the counterfeit coins.



  If 1,2,9 < 3,4,10 the possible pairs are 3,4 - 3,10 - 4,10 - 3,11 - 3,12 - 3,13 - 4,11 - 4,12 - 4,13

   Compare 3,10,X vs 11,12,13


    If 3,10,X > 11,12,13 the possible pairs are 2,3 - 2,10 - 3,10

     Compare 2 vs 3.  If 2 > 3 then 2,10 are the counterfeit coins.  If 2 < 3 then 3,10 are the counterfeit coins.  If 2 = 3 then 2,3 are the counterfeit coins.

    If 3,10,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13

     Compare 11 vs 12.  If 11 > 12 then 2,11 are the counterfeit coins.  If 11 < 12 then 2,12 are the counterfeit coins.  If 11 = 12 then 2,13 are the counterfeit coins.

    If 3,10,X = 11,12,13 the possible pairs are 3,11 - 3,12 - 3,13

     Compare 11 vs 12.  If 11 > 12 then 3,11 are the counterfeit coins.  If 11 < 12 then 3,12 are the counterfeit coins.  If 11 = 12 then 3,13 are the counterfeit coins.



  If 1,2,9 = 3,4,10 the possible pairs are 1,3 - 1,4 - 2,3 - 2,4 - 3,9 - 4,9 - 1,10 - 2,10

   Compare 1,3,9 vs 2,4,X


    If 1,3,9 > 2,4,x the possible pairs are 1,3 - 1,10 - 3,9

     Compare 9 vs 10.  If 9 > 10 then 3,9 are the counterfeit coins.  If 9 < 10 then 1,10 are the counterfeit coins.  If 9 = 10 then 1,3 are the counterfeit coins.

    If 1,3,9 < 2,4,X then possible pairs are 2,4 or 2,10

     Compare 4 vs 10.  If 4 > 10 then 2,4 are the counterfeit coins.  If 4 < 10 then 2,10 are the counterfeit coins.

    If 1,3,9 = 2,4,X the possible pairs are 1,4 - 2,3 - 4,9

     Compare 1 vs 9.  If 1 > 9 then 1,4 are the counterfeit coins.  If 1 < 9 then 4,9 are the counterfeit coins.  If 1 = 9 then 2,3 are the counterfeit coins.


A similar analysis can be made if 1,2,3,4 < 5,6,7,8


If 1,2,3,4 = 5,6,7,8 - either one counterfeit coin is in 1,2,3,4 and the other in 5,6,7,8; both are in 9,10,11,12,13; or all are equal

  Compare 1,2,5,9,10 vs 3,4,6,11,12


   If 1,2,5,9,10 > 3,4,6,11,12 the possible pairs are 1,5 - 1,7 - 1,8 - 2,5 - 2,7 - 2,8 - 9,13 - 10,13

    Compare 1,9,10 vs 5,7,8


     If 1,9,10 > 5,7,8 the possible pairs are 10,13 or 9,13

      Compare 10 vs 9.  If 10 > 9 then 10,13 are the counterfeit coins.  If 10 < 9 then 9,13 are the counterfeit coins.

     If 1,9,10 < 5,7,8 the possible pairs are 2,5 - 2,7 - 2,8

      Compare 5 vs 7.  If 5 > 7 then 2,5 are the counterfeit coins.  If 5 < 7 then 2,7 are the counterfeit coins.  If 5 = 7 then 2,8 are the counterfeit coins.

     If 1,9,10 = 5,7,8 the possible pairs are 1,5 - 1,7 - 1,8

      Compare 5 vs 7.  If 5 > 7 then 1,5 are the counterfeit coins.  If 5 < 7 then 1,7 are the counterfeit coins.  If 5 = 7 then 1,8 are the counterfeit coins. 


   If 1,2,5,9,10 < 3,4,6,11,12 the possible pairs are 1,5 - 1,7 - 1,8 - 2,5 - 2,7 - 2,8 - 9,13 - 10,13

    Compare 3,11,12 vs 6,7,8


     If 3,11,12 > 6,7,8 the possible pairs are 11,13 or 12,13

      Compare 11 vs 12.  If 11 > 12 then 11,13 are the counterfeit coins.  If 11 < 12 then 12,13 are the counterfeit coins.

     If 3,11,12 < 6,7,8 the possible pairs are 4,6 - 4,7 - 4,8

      Compare 6 vs 7.  If 6 > 7 then 4,6 are the counterfeit coins.  If 6 < 7 then 4,7 are the counterfeit coins.  If 6 = 7 then 4,8 are the counterfeit coins.

     If 3,11,12 = 6,7,8 the possible pairs are 3,6 - 3,7 - 3,8

      Compare 6 vs 7.  If 6 > 7 then 3,6 are the counterfeit coins.  If 6 < 7 then 3,7 are the counterfeit coins.  If 6 = 7 then 3,8 are the counterfeit coins.


   If 1,2,5,9,10 = 3,4,6,11,12 the possible pairs are 9,11 - 9,12 - 10,11 - 10,12 - 3,5 - 4,5 - 1,6 - 2,6 - or all are equal

    Compare 1,3,9,11 vs 2,4,10,12


     If 1,3,9,11 > 2,4,10,12 the possible pairs are 9,11 - 3,5 - 1,6

      Compare 1 vs 3.  If 1 > 3 then 1,6 are the counterfeit coins.  If 1 < 3 then 3,5 are the counterfeit coins.  If 1 = 3 then 9,11 are the counterfeit coins.

     If 1,3,9,11 < 2,4,10,12 the possible pairs are 10,12 - 4,5 - 2,6

      Compare 2 vs 4.  If 2 > 4 then 2,6 are the counterfeit coins.  If 2 < 4 then 4,5 are the counterfeit coins.  If 2 = 4 then 10,12 are the counterfeit coins.

     If 1,3,9,11 = 2,4,10,12 the possible pairs are 9,12 or 10,11 or all coins are equal.

      Compare 9 vs 10.  If 9 > 10 then 9,12 are the counterfeit coins.  If 9 < 10 then 10,11 are the counterfeit coins.  If 9 = 10 all coins weigh the same.


Pretty messy to type out and am sure to have made some typos if not poor logic all together.  Do think in theory this puzzle is solvable.  With four questions one can 

discern 3^4 or 81 pieces of information using a balance scale (>,<,= with each balancing) and the total number of possibilities I believe would be 13C2 + 13C0 or 79.

I am in awe. That is brilliant.

[plainglazed]

as shucks Nana, thanks . did make a mistake transcribing from my notes as well as leave out two possibilities but the series of questions does work. here's the corrected solution.

Compare 1,2,3,4 vs 5,6,7,8

If 1,2,3,4 > 5,6,7,8 - there must be two counterfeit coins. either one in 1,2,3,4 and the other in 9,10,11,12,13 or both in 1,2,3,4 with 5,6,7,8 normal (X)
Compare 1,2,9 vs 3,4,10

If 1,2,9 > 3,4,10 the possible pairs are 1,2 - 1,9 - 2,9 - 1,11 - 1,12 - 1,13 - 2,11 - 2,12 - 2,13
Compare 1,9,X vs 11,12,13

If 1,9,X > 11,12,13 the possible pairs are 1,9 - 1,2 - 9,2
Compare 1 vs 9. If 1>9 then 1,2 are the counterfeit coins. If 1<9 then 9,2 are the counterfeit coins. If 1=9 then 1,9 are the counterfeit coins.
If 1,9,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13
Compare 11 vs 12. If 11>12 then 2,11 are the counterfeit coins. If 11<12 then 2,12 are the counterfeit coins. If 11=12 then 2,13 are the counterfeit coins.
If 1,9,X = 11,12,13 the possible pairs are 1,11 - 1,12 - 1,13
Compare 11 vs 12. If 11>12 then 1,11 are the counterfeit coins. If 11<12 then 1,12 are the counterfeit coins. If 11=12 then 1,13 are the counterfeit coins.


If 1,2,9 < 3,4,10 the possible pairs are 3,4 - 3,10 - 4,10 - 3,11 - 3,12 - 3,13 - 4,11 - 4,12 - 4,13
Compare 3,10,X vs 11,12,13

If 3,10,X > 11,12,13 the possible pairs are 2,3 - 2,10 - 3,10
Compare 2 vs 3. If 2 > 3 then 2,10 are the counterfeit coins. If 2 < 3 then 3,10 are the counterfeit coins. If 2 = 3 then 2,3 are the counterfeit coins.
If 3,10,X < 11,12,13 the possible pairs are 2,11 - 2,12 - 2,13
Compare 11 vs 12. If 11 > 12 then 2,11 are the counterfeit coins. If 11 < 12 then 2,12 are the counterfeit coins. If 11 = 12 then 2,13 are the counterfeit coins.
If 3,10,X = 11,12,13 the possible pairs are 3,11 - 3,12 - 3,13
Compare 11 vs 12. If 11 > 12 then 3,11 are the counterfeit coins. If 11 < 12 then 3,12 are the counterfeit coins. If 11 = 12 then 3,13 are the counterfeit coins.


If 1,2,9 = 3,4,10 the possible pairs are 1,3 - 1,4 - 2,3 - 2,4 - 3,9 - 4,9 - 1,10 - 2,10
Compare 1,3,9 vs 2,4,X

If 1,3,9 > 2,4,x the possible pairs are 1,3 - 1,10 - 3,9
Compare 9 vs 10. If 9 > 10 then 3,9 are the counterfeit coins. If 9 < 10 then 1,10 are the counterfeit coins. If 9 = 10 then 1,3 are the counterfeit coins.
If 1,3,9 < 2,4,X then possible pairs are 2,4 or 2,10
Compare 4 vs 10. If 4 > 10 then 2,4 are the counterfeit coins. If 4 < 10 then 2,10 are the counterfeit coins.
If 1,3,9 = 2,4,X the possible pairs are 1,4 - 2,3 - 4,9
Compare 1 vs 9. If 1 > 9 then 1,4 are the counterfeit coins. If 1 < 9 then 4,9 are the counterfeit coins. If 1 = 9 then 2,3 are the counterfeit coins.

The above accounts for 26 possibilities. A similar analysis can be made if 1,2,3,4 < 5,6,7,8 accounting for another 26 possibilities.

If 1,2,3,4 = 5,6,7,8 - either one counterfeit coin is in 1,2,3,4 and the other in 5,6,7,8; both are in 9,10,11,12,13; or all are equal
Compare 1,2,5,9,10 vs 3,4,6,11,12

If 1,2,5,9,10 > 3,4,6,11,12 the possible pairs are 1,5 - 1,7 - 1,8 - 2,5 - 2,7 - 2,8 - 9,13 - 10,13 - 9,10
Compare 1,9,10 vs 5,7,8

If 1,9,10 > 5,7,8 the possible pairs are 10,13 or 9,13
Compare 10 vs 9. If 10 > 9 then 10,13 are the counterfeit coins. If 10 < 9 then 9,13 are the counterfeit coins. If 10 = 9 then 9,10 are the counterfeit coins.
If 1,9,10 < 5,7,8 the possible pairs are 2,5 - 2,7 - 2,8
Compare 5 vs 7. If 5 > 7 then 2,5 are the counterfeit coins. If 5 < 7 then 2,7 are the counterfeit coins. If 5 = 7 then 2,8 are the counterfeit coins.
If 1,9,10 = 5,7,8 the possible pairs are 1,5 - 1,7 - 1,8
Compare 5 vs 7. If 5 > 7 then 1,5 are the counterfeit coins. If 5 < 7 then 1,7 are the counterfeit coins. If 5 = 7 then 1,8 are the counterfeit coins.

If 1,2,5,9,10 < 3,4,6,11,12 the possible pairs are 3,6 - 3,7 - 3,8 - 4,6 - 4,7 - 4,8 - 11,13 - 12,13 - 11,12
Compare 3,11,12 vs 6,7,8

If 3,11,12 > 6,7,8 the possible pairs are 11,13 or 12,13
Compare 11 vs 12. If 11 > 12 then 11,13 are the counterfeit coins. If 11 < 12 then 12,13 are the counterfeit coins. If 11 = 12 then 11,12 are the counterfeit coins.
If 3,11,12 < 6,7,8 the possible pairs are 4,6 - 4,7 - 4,8
Compare 6 vs 7. If 6 > 7 then 4,6 are the counterfeit coins. If 6 < 7 then 4,7 are the counterfeit coins. If 6 = 7 then 4,8 are the counterfeit coins.
If 3,11,12 = 6,7,8 the possible pairs are 3,6 - 3,7 - 3,8
Compare 6 vs 7. If 6 > 7 then 3,6 are the counterfeit coins. If 6 < 7 then 3,7 are the counterfeit coins. If 6 = 7 then 3,8 are the counterfeit coins.

If 1,2,5,9,10 = 3,4,6,11,12 the possible pairs are 9,11 - 9,12 - 10,11 - 10,12 - 3,5 - 4,5 - 1,6 - 2,6 - or all are equal
Compare 1,3,9,11 vs 2,4,10,12

If 1,3,9,11 > 2,4,10,12 the possible pairs are 9,11 - 3,5 - 1,6
Compare 1 vs 3. If 1 > 3 then 1,6 are the counterfeit coins. If 1 < 3 then 3,5 are the counterfeit coins. If 1 = 3 then 9,11 are the counterfeit coins.
If 1,3,9,11 < 2,4,10,12 the possible pairs are 10,12 - 4,5 - 2,6
Compare 2 vs 4. If 2 > 4 then 2,6 are the counterfeit coins. If 2 < 4 then 4,5 are the counterfeit coins. If 2 = 4 then 10,12 are the counterfeit coins.
If 1,3,9,11 = 2,4,10,12 the possible pairs are 9,12 or 10,11 or all coins are equal.
Compare 9 vs 10. If 9 > 10 then 9,12 are the counterfeit coins. If 9 < 10 then 10,11 are the counterfeit coins. If 9 = 10 all coins weigh the same.

And this bit where 1234 = 5678 accounts for the remaining 27 possibilities. (13C2 + 13C0 = 79 total possible combinations)

Label coins 1-13.

Edited July 19, 2011 by plainglazed

[k-man]

I solved this yesterday too, but had to leave, so didn't have time to post my solution. It's slightly different from the one posted by plainglazed, so not to waste good 3 hours of work, here goes...

number all coins 1-13

Step 1. Weigh 1,2,3,4 vs. 5,6,7,8

This leaves 27 possibilities:

1 - no fakes

16 - one fake on each side

10 - 2 fakes among 9-13.

Step 2. Weigh 1,2,5,6,9 vs. 3,7,10,11,12

This leaves 9 possibilities: no fakes or one of 8 fake pairs [1,7],[2,7],[3,5],[3,6],[9,10],[9,11],[9,12]

Step 3. Weigh 1,2,3,4 vs. 7,10,11,12

This leaves 3 possibilities: no fakes or one of 2 fake pairs [1,7],[2,7]

Step 4. Weigh 1 vs. 2

If Equal there are no fakes, if 1>2 then the fake pair is [1,7], if 1<2 then fake pair is [2,7].

This leaves 3 possibilities: [3,5],[3,6],[4,8]

Step 4. Weigh 5 vs. 6

If Equal [4,8] is a fake pair, if 5>6 then the fake pair is [3,5], if 5<6 then fake pair is [3,6].

This leaves 3 possibilities: [9,10],[9,11],[9,12]

Step 4. Weigh 10 vs. 11

If Equal [9,12] is a fake pair, if 10>11 then the fake pair is [9,10], if 10<11 then fake pair is [9,11].

This leaves 9 possibilities: [1,5],[1,6],[1,8],[2,5],[2,6],[2,8],[4,5],[4,6],[9,13]

Step 3. Weigh 1,4,9 vs. 2,5,8

This leaves 3 possibilities: [1,5],[1,8],[4,5]

Step 4. Weigh 5 vs. 8

If Equal then [4,5] is a fake pair, if 5>8 then [1,5] is fake, otherwise [1,8] is fake.

This leaves 3 possibilities: [1,6],[4,6],[9,13]

Step 4. Weigh 1 vs. 4

If Equal then [9,13] is a fake pair, if 1>4 then [1,6] is fake, otherwise [4,6] is fake.

This leaves 3 possibilities: [2,5],[2,6],[2,8]

Step 4. Weigh 5 vs. 6

If Equal then [2,8] is a fake pair, if 5>6 then [2,5] is fake, otherwise [2,6] is fake.

This leaves 9 possible fake pairs: [3,7][3,8],[4,7],[10,11],[10,12],[10,13],[11,12],[11,13],[12,13]

Step 3. Weigh 7,8,10 vs. 11,12,13

This leaves 3 possible fake pairs: [10,11],[10,12],[10,13]

Step 4. Weigh 11 vs. 12

If Equal then [10,13] is a fake pair, if 11>12 then [10,11] is fake, otherwise [10,12] is fake.

This leaves 3 possible fake pairs: [3,7],[3,8],[4,7]

Step 4. Weigh 7 vs. 8

If Equal then [4,7] is a fake pair, if 7>8 then [3,7] is fake, otherwise [3,8] is fake.

This leaves 3 possible fake pairs: [11,12],[11,13],[12][13]

Step 4. Weigh 12 vs. 13

If Equal then [12,13] is a fake pair, if 12>13 then [11,12] is fake, otherwise [11,13] is fake.

This leaves 26 possibilities:

6 - both fakes are on the left side

20 - one fake is on the left and one is among 9-13

Step 2. Weigh 1,9,10 vs. 2,11,12

This leaves 8 possible fake pairs: [1,2],[1,11],[1,12],[2,9],[2,10],[3,4],[3,13],[4,13]

Step 3. Weigh 2,11,12 vs. 4,9,10

This leaves 3 possible fake pairs: [2,9],[2,10],[3,13]

Step 4. Weigh 9 vs. 10

If Equal then [3,13] is a fake pair, if 9>10 then [2,9] is fake, otherwise [2,10] is fake.

This leaves 3 possible fake pairs: [1,2],[1,11],[1,12]

Step 4. Weigh 11 vs. 12

If Equal then [1,2] is a fake pair, if 11>12 then [1,11] is fake, otherwise [1,12] is fake.

This leaves 2 possible fake pairs: [3,4],[4,13]

Step 4. Weigh 3 vs. 4

If Equal then [3,4] is a fake pair, otherwise 4 will be heavier than 3, so [4,13] is fake. 3 cannot be heavier than 4.

This leaves 9 possible fake pairs: [1,3],[1,4],[1,9],[1,10],[1,13],[3,9],[3,10],[4,9],[4,10]

Step 3. Weigh 3,9 vs. 4,10

This leaves 3 possible fake pairs: [1,13],[3,10],[4,9]

Step 4. Weigh 1 vs. 3

If Equal then [4,9] is a fake pair, if 1>3 then [1,13] is fake, otherwise [3,10] is fake.

This leaves 3 possible fake pairs: [1,3],[1,9],[3,9]

Step 4. Weigh 1 vs. 3

If Equal then [1,3] is a fake pair, if 1>3 then [1,9] is fake, otherwise [3,9] is fake.

This leaves 3 possible fake pairs: [1,4][1,10],[4,10]

Step 4. Weigh 1 vs. 4

If Equal then [1,4] is a fake pair, if 1>4 then [1,10] is fake, otherwise [4,10] is fake.

This leaves 9 possible fake pairs: [2,3],[2,4],[2,11],[2,12],[2,13],[3,11],[3,12],[4,11],[4,12]

Step 3. Weigh 3,11 vs. 4,12

This leaves 3 possible fake pairs: [2,13],[3,12],[4,11]

Step 4. Weigh 2 vs. 3

If Equal then [4,11] is a fake pair, if 2>3 then [2,13] is fake, otherwise [3,12] is fake.

This leaves 3 possible fake pairs: [2,3],[2,11],[3,11]

Step 4. Weigh 2 vs. 3

If Equal then [2,3] is a fake pair, if 2>3 then [2,11] is fake, otherwise [3,11] is fake.

This leaves 3 possible fake pairs: [2,4][2,12],[4,12]

Step 4. Weigh 2 vs. 4

If Equal then [2,4] is a fake pair, if 2>4 then [2,12] is fake, otherwise [4,12] is fake.

Follow the same logic as in when Left is heavy, but substitute 1,2,3,4 with 5,6,7,8 respectively

I don't know if these nested spoilers was a good idea. It seemed like a good idea when I started, but now it seems to make it harder to read rather then easier. Oh, well...

[Guest]

Wow guys I have to say I'm a bit speechless. I am amazed at the detail provided here as the logic completely surpasses my own abilities, heh. So thanks a million, I don't believe I could have completed this in any acceptable amount of time.

In any case if anyone is up to it I actually have more. I seriously thought as complicated as the last weighing was that would be the last challenge but sure enough I am presented with another.

1. For this challenge, you are presented with a set of 12 suspect coins. Of these, it

is known that the number of counterfeit coins is either 0, 1 or 2. All of

the counterfeit coins, if any, are known to have come from a single batch

of light coins. You must identify the counterfeit coins, if any, after 4

weighings or less.

2. For this challenge, you have flask A, which holds 19 ounces, and flask B, which

holds 12 ounces. Using only the operations of filling a flask, emptying a

flask, and pouring one flask into another (until either the first is

empty or the second is full), you must manage to put exactly 16 ounces of

water in flask A, leaving flask B empty.

3. At a recent birthday party for a friend of mine, I pointed out the following to my

friend: Eighty-seven years after you were one-third as old as you were one year ago,

you will be four times as old as you were five years after you were one-fourth as old

as you will be eight years from now. What is my friend's age?

Thanks again!

[k-man]

njBLOB,

Welcome to the Den. Please start a new topic for every new riddle/problem. It gets messy when multiple unrelated riddles are posted in the same topic.

[Guest]

Thanks k-man, no problem...