Magnum, Final score Holland 4 Latvia 1.
I think that would work very well. Strange as it seems I have four possible answers.
Original question.
There are three pairs of balls - red, white, and blue. In each pair one ball is a little bit heavier than another one. All the heavy balls weigh the same, and all the light balls weigh the same. Also you have a balance scale.
Now, in just two weighing you have to determine the light and the heavy balls in each pair. How this can be done?
1. Take 2 balls of same color (say blue).
2. Take one ball each of the other 2 colors; and put each one along with a Blue ball on the trays of the balance. (say Blue & White against Blue & Red)
3. There are 2 possibilities:
First:
i) If the balance is equal, then of the Red and the White balls, one is heavy and the other is light.
ii) Keeping track of which Blue ball was paired with which color, weigh the White and the Red balls against each other. Find which one is heavy. (let us say, White turns out to be the heavier one)
iii) Then the Blue weighed with the White is light and the Blue weighed with Red is heavy.
Second:
i) One side is heavier.
ii) The Blue ball on the heavier side is the heavy Blue one.
iii) Put both Blue balls on one side against the White and the Red balls on the other.
iii) If the side with the Blue balls is heavier, then the Red and the White balls are the light ones of their respective colours.
iv) If the other side is heavier, then the Red and the White are both heavy balls.
v) If the balance is equal, the ball that was paired with the heavy Blue ball during the first time is heavy and the other is light.
Example:
Red: R
White: W
Blue: B
1st weighing:
B1R1 Vs B2W1
Equal:
2nd weighing:
R1 Vs W1 (say R1 is heavier, which implies W2 is also heavy)
Then:
B1 is light and B2 is heavy. We know R1 and W2 are heavy therefore R2 and W1 are light.
Say:
B1R1 is heavier, then B1 is the heavy one and B2 the lighter.
2nd weighing:
B1B2 Vs R1W1
If B1B2 is heavier, R1 and W1 are both light, hence R2 and W2 are both heavy.
If R1W1 is heavier, R1 and W1 are both heavy, hence R2 and W2 are light.
If the balance is equal, R1 is heavy and W1 is light, therefore R2 is light and W2 is heavy.
Solution by David Low
Nothing like a long shower to clear out the cobwebs...
Place a red and blue marble on one side, and a red and white marble on the other.
Case 1. The two sides are the same:
Since there are not four heavy marbles or four light marbles, each side must have one light and one heavy marble. Thus the heavy red marble has a light marble with it, and the light red marble has a heavy marble with it.
For the second massing, just compare the two red marbles. The heavy and light red marbles are directly discovered. Moreover, the marble with the heavy red marble in the first massing is now known to be light, and the marble with the light red marble in the first massing is now known to be heavy. The untouched blue and white marbles will respectively be the opposite of their known same-colour partners.
Case 2. One side is heavier.
The heavier side cannot have the light red marble, since such a situation would give at most one heavy marble on the heavier side, and at least one heavy (red) marble on the lighter side, which is impossible. So the heavier side must have the heavy red marble, and the lighter side has the light red marble. The red marbles are discovered.
For the second massing, place the two red marbles on one side, and the one blue and one white marble from the first massing on the other side. If the blue/white side is heavier than the two reds, both marbles are heavy. If the blue/white side is lighter, both are light. If the balance is the same, then one marble is light and the other heavy. The marble that was with the heavy red in the first massing is heavy and the marble with the light red in the first massing is light, since the reverse would have resulted in a tie in the first massing.
As before, the untouched blue and white marbles will respectively be the opposite of their known same-colour partners.
Solution by Gregory Clayborne
This is a great puzzle. The hardest part is giving the solution in simple steps, but here goes...
Let's first label our balls... R1,R2,W1,W2,B1,B2 and then start weighing...
Weighing 1: Lets weigh R1,W1 vs R2,B1
There are two possible results:
The scales will balance or they won't.
If the scales balance: R1,W1 = R2,B1
then we know that there is a Heavy and a Light on each side. We just don't know who's who. We know this because the Reds can't both be Heavy nor can they both be Light. So we have the following possibilities for R1,W1 = R2,B1:
R1(H),W1(L) = R2(L),B1(H) or
R1(L),W1(H) = R2(H),B1(L)
Notice that this makes W1 the opposite of R1 and B1 the opposite of R2 so.....
Reds opposite Whites R1 = W2, R2 = W1
Reds equal Blues R1 = B1, R2 = B2
For the second weighing we just weigh the Reds against each other and the above equations will finish the results.
If the scales don't balance then it gets a little harder.
We know that R1 can't equal R2 and since the scales didn't balance then we know that
which ever side was Heavy has the Heavy Red ball. Don't believe me do you. Alright.
Let's say that the weighing looked like this:
R1,W1 > R2,B1 thus R1,W1 Heavier than R2,B1. If R1 was actually Light (thus making R2 Heavy) then the only values W1 and B1 could have would be Heavy and Light respectively. BUT that would have given us R1(L),W1(H) > R2(H),B1(L). WHICH WOULD HAVE BALANCED (see L,H = H,L) SO R1 has to be Heavy if it's scale went down.
That being proven, let's stick with the assumption that R1(H) and R2(L) just to make the
logic easier to follow...
The first weighing then gives us the results...
W1 and B1 are the same (both Heavy or both Light) OR
W1 opposite B1 with W1 being the same as R1 and B1 being the same as R2.
This leads us to the final weighing...
We just switch R2 and W1 and wind up weighing R1,R2 vs. W1,B1.
If the scales balance then R1(H) R2(L) and W1(H) B1(L)
If the scales don't balance than R1(H) R2(L) and W1 = B1.
But wait, that doesn't tell us if W1 and B1 are Heavy or Light?
Just answer the question. Did the W1,B1 side of the scale go up or down. There's your answer. The final scale would look like one of the following:
R1(H),R2(L) > W1(L),B1(L) OR
R1(H),R2(L) < W1(H),B1(H)
Ta daaaa.
Like I said. This is a great puzzle. Hope you could get through the logic. I really need to draw it out. Gregory
Solution by Shashi
I consider the pair as R1,R2; B1,B2 ; W1,W2
First weighing
R1+W1 against R2+B1
case 1: if it balances then
second weighing,
Weigh R1 and R2 and find out which is heavier, this also tells which of (W1,B1) is heavier.
case 2 if it does not balance
second weighing
weigh R1+R2 against W1+B1
if R1+R2 side goes down then both W1 and B1 are lighter balls of their pairs
if W1+B1 side goes down then both W1 and B1 are heavier balls of their pairs
if it balances
then the ball which was with lighter red ball in the first weighing is lighter i.e., if R1 was lighter then W1 is also lighter
Solution by Marcus Dunstan
6 Balls
White = W1, W2
Red=R1,R2
Blue=B1,B2
First weigh: W1+R1 v W2+B1
If balanced then R1 & B1 must be one HEAVY (H) and one LIGHT (L) (because we know W1 and W2 are one HEAVY and one LIGHT)
If balanced, the second weigh is to swap R1 and B1. ie second weigh is W1+B1 v W2+R1
The side of scales that goes down has 2 HEAVY balls, side of scales that goes up has 2 LIGHT balls.
>From this situation weight of all 6 balls is now known.
If first weigh W1+R1 v W2+B1 is not balanced then you know that the side that goes DOWN must contain the HEAVY white ball eg. W1+R1 v W2+B1
Possible combinations: H+H v L+H, H+H v L+L, H+L v L+L then W1 is HEAVY
or
Possible combinations: L+H v H+H, L+L v H+H, L+L v H+L then W2 is HEAVY
Cannot be combinations: H+L v L+H or L+H v H+L as scales would balance
So at this stage you know the weight of the two white balls only
*** HOWEVER you must also note the colour of the ball on the same side as the LIGHT white ball as this will have bearing depending on the results of the second weigh.
The second weigh would then be W1+W2 v R1+B1
If W1+W2 side goes down then combination must be H+L v L+L or L+H v L+L ie
both R1 and B1are LIGHT
If W1+W2 side goes up then combination must be H+L v H+H or L+H v H+H ie
both R1 and B1 are HEAVY
If W1+W2 balances with R1+B1 then you must have one HEAVY and one LIGHT
ball on the R1+B1 side
The only way this can occur (bearing in mind the results of the first weigh) is if the coloured ball noted above *** was LIGHT
>From this situation weight of all 6 balls is now known.
A horse breeder goes to a horse show with a certain number of horses. The first buyer comes by and purchases half of the horses the breeder brought plus half a horse. The second buyer comes by and purchases half of what remains plus half a horse. The third buyer comes by and purchases half of what remains plus half a horse. The breeder leaves, satisfied that he has sold all the horses he brought.
All three buyers have purchased whole horses, and there is no shared ownership among them.
How many horses did the breeder bring to the show
Laid out neatly in front of you as far as you can see are coins, coins and more coins. "That's a lot of coins," you think to yourself.
Greed sidles up to you and whispers, "Actually, there are an infinite number. And it can all be yours!"
Upon seeing your raised eyebrow, he continues. "The rules are simple. 20 of the coins are heads, the rest are tails. All you have to do is split all the coins into just two groups so that the number of heads is the same in both groups."
Seeing the obvious difficulties, you begin to protest. "But..."
"You can turn over as many coins as you like," he interrupts. "And the groups do not have to have the same number of coins in them."
"Oh ok. Can I start now?"
"Sure, you have one day to complete this task. You can start as soon as I blind you for one day."
"How can I get through it all in a day...! What do you mean blind me...!?"
Although you cannot see him, you are quite certain that he is grinning.
You cannot see. You cannot feel the difference between head nor tail. How can you accomplish this task with certainty in a single day
Farmer John had a problem. There were a group of brigands that had taken all he had... except for three things: his prized wolf, his goat, and a box of cabbages. They were coming after him, to get the rest. These brigands did not like water, so John went to the Blue River, a deep, fast river that no one could swim, and it had no bridges. He always kept a boat there, because he liked to fish, but it was small. So small, in fact, that he and only one of his precious things could be in the boat at the same time.
It sounds simple, right? Ferry one item across at a time, and come back for the others? Well, if John leaves the goat with the cabbages alone on one side of the river the goat will eat the cabbages. If he leaves the wolf and the goat on one side the wolf will eat the goat. If john is there, only he can separate the wolf from the goat and the goat from the cabbage.
How can farmer John keep his possessions safe from the brigands, without losing a single one
If not then you know that on heavier side are heavier red and white balls.
)
So is your answer. I had 17 moves. 
Are you sure you followed the moves with your one eye?