Question

If there are 12 bowling balls, they are all the same colour and size, but the weight of one of them is slightly different in weight only than the other 11.
You have a set of massive rickety old scales, that will fall apart after the 3rd use. If you can weigh the balls in whatever combination you want on the scales (up to 12 at a time), but you only have 3 uses of the scales... how do you find the different ball?
not so much a maths problem... but could use some help please?

Answer 1

It's tricky. I don't remember all the steps.
The first step is: (4 vs 4), 4 left out. The reason is that you must obtain as much information as possible at this step. If it is equal, the different ball is in the 4 left out. looks easy. If there is a different, you have to use some of the 4 left out balls at the second step because you know they are normal.

Answer 2

suppose the first step shows no difference in weight.
you got balls 1,2,...,8 that are equal. the difference lies in balls 9,10,11,12.

put on the left 9,10, on the right 11 and 1!
- nothing happens: compare 12 to 1
- left is heavier: compare 9 vs 10. + make conclusions
- right is heavier: compare 9 vs 10 + make conclusions

Answer 3

so step 1

Answer 4

Buy some new scales. LOL
but @reemii seems to be well on the way to getting it :3

Answer 5

but, are you sure one ball weighs LESS? or is it : "maybe less, maybe more" ?

Answer 6

It is a good to keep track of what ball went through what comparison, so you can use it later as a "potentially heavy".

Answer 7

If we're sure that the faulty ball ways less, it all boils down to (once you've separated the 12 into three sets of four) to find out WHICH set of four balls has the faulty one. This is accomplished after reemii's first weighing.

Answer 8

wouldnt it be whay harder if it was "maybe more, maybe less"...?

Answer 9

@terenzreignz is right. knowing that the ball is lighter makes it easy. you don't need 3 tries.

Answer 10

You do. Once you know which set of four contains the lighter ball, take that set of four, divide it into two sets of two and weigh them again.

The lighter pair would contain the faulty ball, so take that pair, and weigh the individual balls. The lighter ball is the 'winner' :D

Answer 11

still take 3 tries tho, yeah?

Answer 12

yes I saw why. @terenzreignz

Answer 13

I saw the sign :D

Answer 14

weigh 6,6; pick the lighter side

weigh 3,3; pick the lighter side

weigh 1,1; pick the lighter side, or pick the one not weighed

Answer 15

ok... but what if it was "different" instead of lighter, like reemi said...?

Answer 16

that is a problem for your summer vacation..

Answer 17

define the "difference"

Answer 18

Then the first weighing won't tell you squat.

Answer 19

the difference is "maybe lighter, maybe heavier".

Answer 20

it's winter is Aus dude

Answer 21

Well, unless they balance out.

Answer 22

i spose if your trying to pick the one that looks like a pyrimad among spheres ....

Answer 23

hahahaha! @amistre64

Answer 24

I gagged my ice cream ^

Answer 25

ah, if you dont know if its lighter or heavier; then trying to determine the odd ball out

Answer 26

Actually this problem still has a solution, it is possible to determine which one AND if it is heavier or lighter. I explained above the easy case. I remember vaguely the difficult case, which is: (1,2,3,4 vs 5,6,7,8) shows a difference.. You have to make a 5balls vs 5balls comparison with a mix of balls, like (1,2,5,6,10 vs ...) . hm It is not easy to do that by head.

Answer 27

all good reemii, you've given us some hints, i'll have a play with it over my "weekend" holiday ;)

Answer 28

have fun @Jack1

Answer 29

cheers dude, and thanks for the help hey