83 replies to this topic

[Jkyle1980]

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.

Spoiler for Here's a very slightly modified approach

Use Jkyle1980's groups of nine coins: A, B, C.

[1] A = B: C has a heavy or light coin [18 cases]
[2] A < B: A has a light coin or B has a heavy coin [18 cases]
[3] A > B: A has a heavy coin or B has a light coin [18 cases]

[1] Split C into groups of three coins: C1, C2, C3 and weigh C1 vs. C2.

C1 = C2: C3 has a heavy or light coin [6 cases] Name the C3 coins c3a, c3b, c3c.
c3a = c3b: if c3a </> c3c, then c3c is the heavy / light coin.
c3a < c3b: if c3a </= c3c, then c3a is light / c3b is heavy.
c3a > c3b: if c3a =/> c3c, then c3b is light / c3a is heavy.

C1 < C2: C1 has a light coin or C2 has a heavy coin [6 cases]
C1 = C3: [C2>C1=C3] C2 has a heavy coin.
Procedure 1:
Name the coins c2a, c2b, c2c.
c2a = c2b: c2c is heavy
c2a < c2b: c2b is heavy
c2a > c2b: c2a is heavy
C1 < C3: [C2>C1<C3] C1 has a light coin. Apply Procedure 1 to determine the light coin.
C1 > C3: [C3>C1>C2] Impossible.

C1 > C2: C1 has a heavy coin or C2 has a light coin [6 cases]
C1 = C3: [C1=C3>C2] C2 has a light coin. Apply Procedure 1 to determine the light coin.
C1 < C3: [C2<C1<C3] C1 Impossible.
C1 > C3: [C2<C1>C3] C1 has a heavy coin. Apply Procedure 1 to determine the heavy coin.

[2] Weigh A vs. C

A = C: [A=C<B] B has a heavy coin [9 cases]
Procedure 2:
Split the B coins into groups of three: B1, B2, B3
B1 = B2: B3 has a heavy coin. Apply procedure 1 to determine which.
B1 < B2: B2 has a heavy coin. Apply procedure 1 to determine which.
B1 > B2: B1 has a heavy coin. Apply procedure 1 to determine which.
A < C: [B>A<C] A has a light coin [9 cases] Apply Procedure 2 to determine which.
A > C: [B>A>C] Impossible

[3] Weigh A vs. C

A = C: [A=C>B] B has a heavy coin [9 cases] Apply Procedure 2 to determine which.
A < C: [B<A<C] Impossible
A > C: [B<A>C] A has a heavy coin [9 cases] Apply Procedure 2 to determine which.

So, like I said, 4 right?

[PDR]

I think I've got it in 4 measurements:

Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.

silly me - I was so excited to get a better answer than 9 weighings , that I missed the simplicity of just 4. Good job jkyle1980...

[roolstar]

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?

In fact the answer is 2!!!!

You need a minimun of 2 weighs to find the bad coin IF YOU ARE LUCKY!

1- You take 2 coins, you weight them and you get an unbalanced situation.
2- You take one of them and you compare to an outside coin and you still find an unbalanced situstion ==> The bad coin is X and its 9g or 11g

So the minimum ways needed are 2.

Since we already know this type of problems, it's easy to assume what was meant by the OP.

However I'm sure that what was meant originally was:
What is the minimum number of weighs needed to guarantee finding the "Bad" coin and its weight are...

Spoiler for and...

4

Well, if we knew the real weight of the "bad" coin (9 or 11) we would be able to do it in 3 weighings for 27 coins (discussed previously on this forum)

But in our case here, all we need is ONE extra weighing when we get to an unbalanced situation in any of our weighings (Let's say with groups A & B):

We take one of the two unbalanced groups (A) and compare it to a group of the same number of coins from the rest of the coins (ALL GOOD COINS).
This will tell us which group amongst A & B contained the bad coin AND the real weight of that coin (heavier or lighter). Then we proceed with the rest of the steps...

[Sean-in-Oz]

Folks,

This one took a little testing before I came up with the correct answer . . . FOUR tests.

Some of the flaws I've seen are getting to the final nine in the first weighing. If Group A is the same as Group B, then Group C has the heavy OR light coin. If B was heavier (just to pick one case) however, we need a second test.

Test two would then compare Group B against Group C. If these are the same, then A contains a Light Coin; if B & C are equal, then B contains a heavy coin.

The next two rounds would play out differently depending upon what we learned in Test 1, but perhaps it is easier to show my decision making "system."

###

Let's assume you know that Coin #6 is a light coin, but I know nothing.

Test 1
1-9 vs. 10-18, Group A is lighter.

Test 2
10-18 vs. 19-27, these are the same, therefore I now know two facts. The odd coin is light, and the odd coin is in Group A. Remember if B had been heavier in test two, the odd coin would have been a heavy one in Group B -- You knew that the odd coin was light and only now do I know it as well.

Test 3
1-3 vs. 4-6. Group 2 is Lighter. Now since we know that the coin had to be light, we are down to a final three. If Group 1 had been lighter then 1-3 would be the final group, if they had been the same then Group 3 would be the final group.

Test 4
4 vs. 5. They are the same, so I now know what you knew all along, Coin #6 is light.

Could you get the answer in fewer tests? A qualified NO.

Here is another variation. Assume coin 27 is heavy . . .

Test 1
1-9 vs. 10-18, weigh the same so Group C has the odd coin (lighter? heavier? we don't yet know).

Test 2
19-21 vs. 22-24, weigh the same so Group 3 has the odd coin (lighter? heavier? we don't yet know).

Test 3
25 vs. 26, weigh the same so coin #27 is the odd coin (lighter? heavier? we STILL don't know).

Test 4
26 vs. 27, Coin 27 is the heavy one.

Still Four tests. Basically you need three tests to definitively determine the odd coin, and a fourth test to determine whether the odd coin is heavy or light. The order in which you learn these facts is irrelevant. Note in the first example we learned two of the critical facts after Test 2, and then learned one more fact in each Test 3 and Test 4. In the second example we learned one fact in each of the four tests. The order didn't matter.

So the reason I say a "Qualified No," is that a previous respondent stated that he could be lucky with just two Tests. This is correct. Assume again that coin 27 is the heaviest, and he is lucky enough to choose the following tests.

Test 1
26 vs. 27, Coin 27 is heavIER (either 27 could be the heaviest coin, or 26 could be the lightest)

Test 2
25 vs. 27, Coin 27 is heavier, so he can then conclude that Coin #27 is the HEAVIEST coin.

But if he plays the wild guessing game, and I use my "system," we can easily conclude that analytical thinking will win out over time.

I will always KNOW the relative weight of the odd coin in FOUR tests. He may GUESS it in TWO, but if he is really unlucky he may require as many as FOURTEEN tests (to prove the worse case, again 27 is heavy):

Test 1: 1 vs. 2 Same
Test 2: 3 vs. 4 Same
Test 3: 5 vs. 6 Same
Test 4: 7 vs. 8 Same
Test 5: 9 vs. 10 Same
Test 6: 11 vs. 12 Same
Test 7: 13 vs. 14 Same
Test 8: 15 vs. 16 Same
Test 9: 17 vs. 18 Same
Test 10: 19 vs. 20 Same
Test 11: 21 vs. 22 Same
Test 12: 23 vs. 24 Same
Test 13: 25 vs. 26 Same
Test 14: 1 vs. 27, 27 is heavier, and therefore the heaviest.


His is not exactly a winning system!

[brhan]

I think I've got it in 4 measurements:

Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.

Nice one ...

[Nebula]

2 tests..
Divide into 3 piles
Compare 2 piles with the third?
Two of the piles has equal weight, when you find them you only need to compare one of them with the last to figure out if the coin is heavier or lighter - let one pile stay on the scale.
1st try: 2 equal piles
2nd try: the scale goes up or down
OR
1st try: the scale goes up or down
2nd try: the scale has to go same as 1st

- Not sure if I’ve explained it properly, my English is kinda bad <_<

[MrB]

I suppose this can be done in a simpler way than discussed above!

Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done-

1. Make 2 stacks of 13 coins each and put on A&B.
Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice!
BUT, If Pan B is heavy-

2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again.
IF u'r lucky then A&B would weigh equal & the left over coin is heavier one!
BUT If either side of pan is heavy (lets take A this time!)

3. Take the 6 coins from A, divide into 3 coins each & place on A&B again!
No luck here buddy ! One of either side of pan would be heavy (lets take B this time!)

4. Take the 3 coins, keep one coin on each side of A&B.
WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!!

I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations.

[MrB]

I suppose this can be done in a simpler way than discussed above!

Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done-

1. Make 2 stacks of 13 coins each and put on A&B.
Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice!
BUT, If Pan B is heavy-

2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again.
IF u'r lucky then A&B would weigh equal & the left over coin is heavier one!
BUT If either side of pan is heavy (lets take A this time!)

3. Take the 6 coins from A, divide into 3 coins each & place on A&B again!
No luck here buddy ! One of either side of pan would be heavy (lets take B this time!)

4. Take the 3 coins, keep one coin on each side of A&B.
WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!!

I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations.

Jus an addition is dat this is the solution if v consider the coin to be heavy one. Same can be applied if the coin is lighter one!!!

[rookie1ja]

I suppose this can be done in a simpler way than discussed above!

Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done-

1. Make 2 stacks of 13 coins each and put on A&B.
Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice!
BUT, If Pan B is heavy-

2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again.
IF u'r lucky then A&B would weigh equal & the left over coin is heavier one!
BUT If either side of pan is heavy (lets take A this time!)

3. Take the 6 coins from A, divide into 3 coins each & place on A&B again!
No luck here buddy ! One of either side of pan would be heavy (lets take B this time!)

4. Take the 3 coins, keep one coin on each side of A&B.
WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!!

I suppose this is a much simpler solution than the ones dealing in making 3 stacks & comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations.

the problem is that it is not clear whether the different coin is heavier or lighter

[brain not brawn]

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?

easy.
you ask the coin if it weighed itself lately and to tell you its weight