Try out the ODDBALL logic test?

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Alan McDougall

Alan McDougall
Registered Senior Member
Lets us have a little fun, try this test of logic it is great fun

If you know the solution please refrain from posting unlit the other guys have had a go at it

Alan

12 Steel Balls Logic Test
18 Mar. 07 at 1:13 am

While checking my web statistics today I noticed that the top search keyphrase I’m getting traffic from is “Mind Puzzles”, so maybe it’s time for a new one today.

The 12 steel balls puzzle is known as a logic test, a simple puzzle that doesn’t require any extraordinary math knowledge, but instead it tests the mind logic and the ability to analyze problems, that’s why it’s recommended for developers to test their logic and ability to see all possible solutions of a problem and choose the right route to go through.

The Puzzle
As you can tell from its name, the puzzle is about 12 steel balls, identical in almost everything, color, size and shape, but as for weight they are all identical except for one, which is different without knowing whether it’s a heavier or lighter ball, Now we have a balance scale that we can use for only three weighings to identify the different ball and whether it’s a heavier or a lighter ball.

It may sound simple but it’s not as simple as it sounds, you can find more about this puzzle, its history and some hints and tips to solve on Paul Harkins’s website, I created a PDF version of the same file for non-windows user as well
 
Solution:

You can go with a lottery attempt, namely put 5 balls on each side, if they weigh the same, you scrub 10 from being tested, making the conclusion in the second weigh-in.

However that assumes you get the heavy one in the two left over.

Otherwise:
  • Firstly Put 6 on each side, remove the 6 from the light side.
  • Split the 6 into 3 on each side, weigh again. Remove the 3 from the light side.
  • Put 1 on each side and have one left over not on the scales. If the scales tilt one way or another than you can find the heaviest, if they are balanced then the ball thats not being weighed is the heaviest

[edit]These were quick and nasty (and inaccurate) solutions, the solution is on my post further down the thread.
 
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Won't work.
What if on the second balancing, the balls balance.
You will know then that there is a heavier ball in the other six but you have only one weighing left.
 
I think I've got it.

Weigh 4 of the balls against another 4.

If they balance the odd ball is in the other 4.
Call the balls that balance "Standard balls".

Weigh 3 of the remaining balls against 3 Standard balls.
If they balance, the remaining Ball is the odd ball and can be weighed against a standard ball to see if it is heavier or lighter.

If there is an imbalance, call the lighter set of 3 balls "Set A" and the heavier set " Set B". Call the remaining balls "Standard Balls"

Weigh 3 Balls from Set A and one ball from set B against 4 standard balls. If they balance, the ball is heavy and is one of the other 3 balls from Set B.

If they are lighter, the ball is lighter, and one of the 3 Balls from Set A.
Weigh any two of these to find which is the odd ball. If they balance, then it is the other one.

If they are heavier, the Ball is the Ball which was chosen from set B and is heavier.

This probably has a really simple mathematical answer.
 
Captain Kremmen said:
Won't work.
What if on the second balancing, the balls balance.
You will know then that there is a heavier ball in the other six but you have only one weighing left.
Click to expand...

I apologise I stated it was the method to find the Heaviest ball, not identify a ball that could be heavy or light ;)

To be honest most of the time in the real world we work with the simplest and speediest equation over the most logical. It's apart of what makes us what we are, Guestimations.
 
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Query Kremmen, did you actually use 12 objects on a desktop to work it out, or sit there and attempt to think it out?
 
Stryder said:
Query Kremmen, did you actually use 12 objects on a desktop to work it out, or sit there and attempt to think it out?
Click to expand...

Thought it out, using pen and paper.
If my answer is wrong, I give up. It made my head go hot.

Mathematically, these problems usually use binary numbers.
 
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Wait a second. I was thinking three weighings implies all the balls dont fit on the scale at six per side. If six balls per side fit on the scale then it can be done in one weighing. Very simple, no paper, no marked balls. I dont think a balance scale matter either because it can be done on any scale with just adding the 12 balls.
 
Does the balance scale have markings? It needs to have markings otherwise yu would not be able to tell which of the two balls is either lighter or heavier. Unless you can eye it up but that is a little too hard to do and some people may not have good enough vision to see the difference if it is not very obvious.

Tbh, after two balls you already know the expected weight of the balls....IF the scale has markings. And if the first two balls are not the different wight ball.
 
John99 said:
Does the balance scale have markings? It needs to have markings otherwise yu would not be able to tell which of the two balls is either lighter or heavier.
Click to expand...
You do know what a balance scale is don't you?

balance_scale.jpg

:rolleyes:

If six balls per side fit on the scale then it can be done in one weighing.
Click to expand...
Rubbish.
 
Dywyddyr said:
You do know what a balance scale is don't you?

balance_scale.jpg

:rolleyes:


Rubbish.
Click to expand...

Some balance scales have markings to see how far the scale goes down on one side. Very simple test. The reason is because all the other balls will weigh it down to the same exact level. If not it is a matter of being able to actually see the difference and if it is slight then it is just a matter of the limitation of the person being able to even differentiate the difference. Otherwise the test is not exact anyway.
 
John99 said:
Some balance scales have markings to see how far the scale goes down on one side. Very simple test. The reason is because all the other balls will weigh it down to the same exact level.
Click to expand...
What do you mean "all the other balls"? You said one weighing. Which means all the balls are on there together.

Otherwise the test is not exact anyway.
Click to expand...
What do you mean "exact"? The problem is to distinguish which ball is different in weight. Once you have that single different ball, and whether it's heavier or lighter, then you can't get any more exact as an answer.
 
Dywyddyr said:
Which would constitute TWELVE weighings, not one.
Get a brain. :rolleyes:
Click to expand...

It would be six weighings. He didnt say they could not be weighed one at a time. He said three weighings to weigh 12 balls. Once twelve balls are weighed (one per side) this constitutes one weighing.


As the question is now my answer is correct. So that cannot change. The only way is to change the puzzle.
 
John99 said:
It would be six weighings. He didnt say they could not be weighed one at a time. He said three weighings to weigh 12 balls. Once twelve balls are weighed this constitutes one weighing.
Click to expand...
If you have 12 balls and they are put on one at a time then that's 12 weighings.
The question stipulated 3 weighings, ONLY.
only three weighings
Click to expand...
Get a clue. Or stop posting.
 
Dywyddyr said:
If you have 12 balls and they are put on one at a time then that's 12 weighings.
The question stipulated 3 weighings, ONLY.

Get a clue. Or stop posting.
Click to expand...

He said 3 weighings to weigh twelve balls. Let someone else decide if the answer is wrong.:mad:
 
Meh... I think I know ;)
(Solution in box: You'll have to mouse over to see it)

Code: 
[color=white]
Simplest method was to create a Matrix array;

........PanA PanB Leftover
---------------------------
step 1: abcd efgh ijkl
step 2: ahfg kdli cejb
step 3: kgdj ible acfh

The letters represent each ball, weigh PanA with PanB each step and note the scale positions.
On the last scale test, it should be possible to deduce which ball is lighter or heavier,
since it's position during weighing changes.[/color]

[edit]Fixed the solution since I put in the wrong one lol
 
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