[Guest]

Having 6 balls, equal in shape, size and colour.

Three balls heavier than the other three, but both sets of three balls weighing the same.

How would you identify the all heavier three balls if you could use a pair of scales only three times?

sol.bmp

[Guest]

If 3 balls are heavier, for the sake of argument lets say 4 lbs each, and the other 3 balls weigh, lets say 2 lbs each, it is impossible to have 2 sets of three that weigh the same amount. The total is 18 lbs which means each set of 3 must weigh 9 lbs which is impossible using 4 lb & 2 lb balls since 9 is not equally divisible by either 4 or 2. You can plug any weights you want in and you get the same result, it can't happen.

Having 6 balls, equal in shape, size and colour.

Three balls heavier than the other three, but both sets of three balls weighing the same.

How would you identify the all heavier three balls if you could use a pair of scales only three times?

sol.bmp

Edited December 17, 2008 by HokieKen

[Guest]

If 3 balls are heavier, for the sake of argument lets say 4 lbs each, and the other 3 balls weigh, lets say 2 lbs each, it is impossible to have 2 sets of three that weigh the same amount. The total is 18 lbs which means each set of 3 must weigh 9 lbs which is impossible using 4 lb & 2 lb balls since 9 is not equally divisible by either 4 or 2. You can plug any weights you want in and you get the same result, it can't happen.

I meant, the lighter balls each weigh the same, equally, the heavier balls each weigh the same.

Sorry, I should have been more clear.

[Guest]

Having 6 balls, equal in shape, size and colour.

Three balls heavier than the other three, but both sets of three balls weighing the same.

How would you identify the all heavier three balls if you could use a pair of scales only three times?

sol.bmp

By the way, if none of you understand my 'spoiler for solution', I can clarify it. It is a lot more work though, and the picture seemed clear.

[Guest]

This is the way I interpreted the question.

You have six balls. All are the same size and colour.

Three balls weigh slightly more than the other three. Eg, three weigh 1.1g and the other three weigh 1.0g.

You have a set of scales (the type with two trays which can show which side is heavier).

How can you identify which balls are heavier using the scale only three times?

Put all six balls on the scales. Naturally, one side will be guaranteed to be heavier than the other.

Now re-weigh the lighter side by putting one marble in each side while leaving the other off. If their weight is different, you'll know that the heavier ball weighs 1.1g and the other two weigh 1.0g. Now weigh the other side the same way. If the two balls have different weights, the lighter ball weighs 1.0g and the other two weigh 1.1g. If they weigh the same, they both weigh 1.1g and the third weighs 1.0g.

There is a problem if the second time you use the scales, both balls are the same weight. You can easily determine that these two weigh 1.0g but you do not know what the third ball weighs. If you continue normally and the next two balls have different weights, you'll know that that ball weighs 1.1g. If not, it's a 50/50 guess.

However, this situation has a 20% chance of occurring. Given that you have a 50/50 chance, using this method you'll get it right 90% of the time.

That's the best I can do. Can anyone do better?

EDIT: No, the picture makes no sense as we have no idea what the numbers mean.

Edited December 18, 2008 by Luke Warmwater

[Guest]

EDIT: No, the picture makes no sense as we have no idea what the numbers mean.

The numbers represent the six different balls, the three numbers at the end of every 'branch' (see it as a kind of chance tree) on the right represent the heavier balls

Also, every 'fraction' is a setup for the pair of scales (for example, you begin by weighing balls 1 and 2 against balls 3 and 4)

Haha hope it makes sense now

Edited December 18, 2008 by lunkkun

[Guest]

Okay, it makes sense now.

Here's an easier one for you:

You have 9 balls, of which eight have an equal weight and one weighs slightly more.

You can only use the scales twice.

Edited December 18, 2008 by Luke Warmwater

[Guest]

Okay, it makes sense now.

Here's an easier one for you:

You have 9 balls, of which eight have an equal weight and one weighs slightly more.

You can only use the scales twice.

Yeah, I solved that one allready.

[Guest]

Having 6 balls, equal in shape, size and colour.

Three balls heavier than the other three, but both sets of three balls weighing the same.

How would you identify the all heavier three balls if you could use a pair of scales only three times?

sol.bmp

this is too easy

take any 2 on scale, heavier type A, lighter type B.

then 2 more, etc...

NOW try with 6 balls, all look the same and only ONE is heavier AND u can only use the scale TWICE!!!

[Guest]

Okay, it makes sense now.

Here's an easier one for you:

You have 9 balls, of which eight have an equal weight and one weighs slightly more.

You can only use the scales twice.

too easy

u take 6, 3 on each scale if both weight same

the ball is in remaining 3, then u take two of them, if same the remaining one.

[Guest]

this is too easy

take any 2 on scale, heavier type A, lighter type B.

then 2 more, etc...

The solution is a bit harder than that, if you take two random balls, they might just weigh the same, and you wo'nt get any further. With this method, three times isn't enough most of the time.

Edited December 18, 2008 by lunkkun

[Guest]

Your picture is nice looking but it's too hard for anyone (at least me) to check its correctness.

Really, I tried to check it, but couldn't managed.

Of course you checked it before post, and I believe its correctness.

My solution is more simple and no need to check it:

Let the balls be 1,2,3,4,5,6

First try: 12 - 34

if 12 is heavier then

eather 1 and 2 are heavy, and 3 or 4 is heavy or

eather 1 or 2 is heavy, and 3 and 4 are light.

Second try: 1 - 2

if they're equal then 1 and 2 are heavy, 3 or 4 is heavy.

A last try 3 - 4 will reveal it. Obviously 5 and 6 are light.

If 1 is heavier than 2. Then 1 is heavy, 2,3,4 are light, 5,6 are heavy. No need to 3.try.

If in the first try 12 = 34, then two tries between 1-3 and then 5-6 will get it.

[bonanova]

the picture makes no sense as we have no idea what the numbers mean.

Numbers are balls being weighed.

Each outcome = lighter, equal, heavier = points to the next weighing.

After three weighings the light and heavy balls are shown.

[Guest]

Your picture is nice looking but it's too hard for anyone (at least me) to check its correctness.

Really, I tried to check it, but couldn't managed.

Of course you checked it before post, and I believe its correctness.

My solution is more simple and no need to check it:

Let the balls be 1,2,3,4,5,6

First try: 12 - 34

if 12 is heavier then

eather 1 and 2 are heavy, and 3 or 4 is heavy or

eather 1 or 2 is heavy, and 3 and 4 are light.

Second try: 1 - 2

if they're equal then 1 and 2 are heavy, 3 or 4 is heavy.

A last try 3 - 4 will reveal it. Obviously 5 and 6 are light.

If 1 is heavier than 2. Then 1 is heavy, 2,3,4 are light, 5,6 are heavy. No need to 3.try.

If in the first try 12 = 34, then two tries between 1-3 and then 5-6 will get it.

If 1 & 2 are both heavy and 3 & 4 are both light, then how to finish in 3 weighings.

[Guest]

If 1 & 2 are both heavy and 3 & 4 are both light, then how to finish in 3 weighings.

Sorry, I missed it, you're right.

I suppose the OP's picture is the unique solution. I admire him.