can u solve this problem

[KingDingaling]

Six balls confront you: two red, two green, and two blue. In each color pair you know that one ball is heavier than the other. You also know that all three of the heavier balls weigh the same, as do all three of the lighter balls. The six balls (call them R1, R2, G1, G2, B1, and B2) are otherwise indistinguishable. You have only a balance scale. With no more than two weighings on the scale, how can one identify the heavier and the ligheter balls in all three pairs?

Anybody got a clue?

[TheDfromGC]

all u would do is take both balls from the same colored pair and put them on different sides of the scale, the heavier one would make the scale goes down.....seems pretty simple unless im missing something.....

[Psychotron]

Take R1 and G1 put it on Scale. If G1 is heavier, take it off, put B1 on. If B1 is heavier take R1 off and put G1 back on, if G1 is heavier, it goes R1 B1 G1

[KingDingaling]

what if r1 and g1 weigh the same?

[SplinterCell]

Brain.....hurting.....ahhhhh

[KingDingaling]

anyone???

[mushroomstampr]

he said balls....hehehe

[symatech]

what if r1 and g1 weigh the same?

then they are either both the lighter or both the heavier. replace an arbitrary one (say g1) with another ball say (g2) if g2 is heavier than r1 then you know g1 and r1 are light. you can figure the rest out from there.

but the first answer you got in this thread is the easiest.

[TheDfromGC]

but the first answer you got in this thread is the easiest.

seems like im missing something because my answer was so simple...

i dont know about everybody else but if 3 sets of balls confront me i wouldnt want to stick around to see whos weigh more...

[RATM]

this sux, i cant figure it out. im a retard

[damiongage]

all u would do is take both balls from the same colored pair and put them on different sides of the scale, the heavier one would make the scale goes down.....seems pretty simple unless im missing something.....

you only get 2 weighings...your answer would require three

[damiongage]

Take R1 and G1 put it on Scale. If G1 is heavier, take it off, put B1 on. If B1 is heavier take R1 off and put G1 back on, if G1 is heavier, it goes R1 B1 G1

two many weighings...you only get 2

[J.S.N.]

feel each ball b4 you put them on the scale so you know which ones are heavier. (j/k)

this is tuff. let's say that the weight of each ball is either 1 or 2. with 1 of each color on each side the weight can either be 3, 4, 5, or 6.



it will either be

1r,1b,2g (4) vs. 2r,2b,1g (5)
1r,1b,1g (3) vs. 2r,2b,2g (6)
1r,2b,1g (4) vs. 2r,1b,2g (5)
2r,1b,1g (4) vs. 1r,2b,2g (5)

now switch the reds and greens to the opposite sides and weight again

this should give you the idea but if you need more help i'll figure out the rest for you i'm too lazy right now.

[TheDfromGC]

do the balls talk?? if they do just ask them

[J.S.N.]

oh wait here i figured it out.

take and put just the reds on. let's say it goes right.

now put one both green and blue on the left and green on the right.

if it balances then the right is red (of course) heavy, green light, blue light.

if it tips left both blue and green on left are heavy, and red on right is heavy.

if it tips right red and green are heavy and the blue on the floor is heavy.

[gixxer600]

do what I do, just jiggle them!! lol

[TheNatural3456]

lol just use your eyes....it doesnt say u cant open em....righjt?

[J.S.N.]

hey no fair that took like .5 hour to solve and hurt my brain and the guy never even saw the answer!!!!!!!!!!!!!!!!!

[SwoleCat]

Balls are overated!

~SC~

[KingDingaling]

oh wait here i figured it out.

take and put just the reds on. let's say it goes right.

now put one both green and blue on the left and green on the right.

if it balances then the right is red (of course) heavy, green light, blue light.

if it tips left both blue and green on left are heavy, and red on right is heavy.

if it tips right red and green are heavy and the blue on the floor is heavy.

actually this isn't right
Ur longer answer isn't right either, unless i'm reading it wrong. let me try and rewrite wut ur saying using my little visual system.

r1 r2
-------

R1 = red ball number 1
R2 = red ball number 2
---- = the scale

So, if it tips right then you know:

LR1 HR2
------------

L = light
H = Heavy

Your next step confuses the **** outta me, but i think ur saying:

G1/B1 G2/HR2
-----------------

to balance u would need:

HG1/LB1 LG2/HR2
---------------------

to tip left would be:

HG1/HB1 LG2/HR2
----------------------

To tip right u need:

LG1/LB1 HG2/HR2

OR

LG1/HB1 HG2/HR2

So that doesn't solve the problem, cause we don't know if blue is heavy or light

[J.S.N.]

i prolly didn't explain it well enough

here i'll make it more visual:

you know which red is heavier on your first weigh-in, and then you put a green on each side, and a blue on the light red side; the options are:

1 2
1 2
1

1 2
2 1
1

1 2
2 1
2


1 2
1 2
2

for the first one, the heavy red side would still tip.

for the second one, it would balance toward red light

for the third one, it would tip toward the light red side.

for the fourth one, it would balance.

then you know the weight of the other blue ball that is one the groud too.

oh **** my bad i did screw it up, if it tips toward light red you don't know the weight **** i'll keep at it sorry.

[*Narkissos*]

arrrrrrrrrrg...**** you all to hell

[J.S.N.]

here i foud a solution online:


On a Christmas tree there were two blue, two red and two white balls. All seemed the same, however in each colour pair there was one ball was heavier. All three lighter balls were the same weight, just like all three heavier balls. Using a pair of scales twice, identify the lighter balls.




Lay one red and one white ball on left pan and one blue and the other white ball on the right pan. If there is equilibrium, then it is clear that there is one heavier and one lighter ball on each side. That’s why comparing white balls is enough to learn everything.
However, if at first weighing one side is heavier, then there must be a heavier white ball on that side. The next reasonable step is to compare the already weighed red ball and yet not weighed blue ball. After that, the character of each ball is clear, isn’t it?