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# Weighing VII.

28 replies to this topic

### #11 Martini

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Posted 28 August 2007 - 04:54 AM

If you are lucky, one weighing is all you'll need. Place one ball on each opposing scale tray and if it tilts, you found the heavy ball. Then discard the other 25.

The puzzle does not specify being able to routinely perform this task.

Yeah, I know, you'll say it is implied.

Yeah, it's implied. But you're right; the riddle should state "what is the minimum number of weighings needed to guarantee a confirmation of which ball is the heavy one?".

It should also be stated that the other 26 balls weigh the same or the minimum number of weighings won't be the same as what the solution gives as an answer.
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### #12 rookie1ja

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Posted 28 August 2007 - 09:24 AM

aaaaaah, those implications ... still leaving space for discussion ... I have added an edit to the riddle
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### #13 Fysh

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Posted 04 September 2007 - 02:40 PM

unfortunately, the given solution doesn't meet the criteria of the problem. Because it asks for a minimum, three is not correct. Logically, you would be able to find it with one weighing. This requires quite a bit of luck, but it does meet the criteria of the problem.
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### #14 rookie1ja

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Posted 05 September 2007 - 08:50 AM

unfortunately, the given solution doesn't meet the criteria of the problem. Because it asks for a minimum, three is not correct. Logically, you would be able to find it with one weighing. This requires quite a bit of luck, but it does meet the criteria of the problem.

as per my latest edit:
What is the minimum number of weighings needed to guarantee a confirmation (always!) of which ball is the heavy one? Of course, the other 26 balls weigh the same.

Are there still any holes? Yes - I held one tennis ball in my hand a bit longer and the sweat made it a bit heavier, there was a chewing gum attached at the bottom of one weighing machine plate (which I forgot to mention) etc.
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### #15 connie

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Posted 29 September 2007 - 10:17 PM

A pair of scales is like a pair of pants. One such as the lady justice statue holds. Anyway I come up with four weighings to guarantee success. Divide balls in two piles of thirteen and set the extra aside. Assuming you didn't luck out, one side will weigh more. Divide the heavier side into two piles of six and set the extra aside. Weigh the two piles of six. Again assuming your still not lucking out, one side will be heavier. Divide the heavier six into two piles of three & weigh them. The heavier side of three balls will contain the heavy ball so set one on each side and set the extra aside. If the scales balance the extra is it. If they don't balance the heavier side is it.
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### #16 tweak4

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Posted 01 October 2007 - 06:23 PM

Ever wonder why people bother posting the wrong answer even after the correct answer has been given in the original post, and even confirmed (often multiple times) by later posters?

The correct answer is 3. No more. No less. 3.
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### #17 Alchameth

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Posted 13 December 2007 - 11:26 PM

I know this is a bit old, but given the question... wouldn't the answer be 0?

If one ball is heavier it would bounce differently then all the others. So hold them at equal heights and drop them 2 at a time and if they bounce back to the same height then discard them.

Maybe 1 if you decided to confirm the "bounce test" results via the scale.
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### #18 mhartnag

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Posted 26 January 2008 - 06:00 PM

The question does not specifically say how many times to absolutely identify it. You can identify the heavier ball with one weighing if you weigh 13 balls on each scale and the mass is equal. It will mean you are holding the heavy ball. Therefore it is possible to identify the heavy ball in one weighing according to the wording of the question.
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### #19 rookie1ja

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Posted 26 January 2008 - 06:29 PM

The question does not specifically say how many times to absolutely identify it. You can identify the heavier ball with one weighing if you weigh 13 balls on each scale and the mass is equal. It will mean you are holding the heavy ball. Therefore it is possible to identify the heavy ball in one weighing according to the wording of the question.

the puzzle was edited a loooooong time ago ... pls read the puzzle thoroughly

Edit: What is the minimum number of weighings needed to guarantee a confirmation (always!) of which ball is the heavy one? Of course, the other 26 balls weigh the same.

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### #20 roolstar

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Posted 26 January 2008 - 08:02 PM

The fact that we only need 1 weighing for 3 balls means that:

we need only 2 weighings for 9 balls...

we need only 3 weighings for 27 balls...

we need only 4 weighings for 81 balls...

we need only 5 weighings for 243 balls...

and 6 weighing for 729 balls...

and 7 weighings for 2187 balls...

And the general formula is weighs needed are x with x the solution of 3x = A!
Of course A is the number of balls or the first power of 3 superior to the number of balls

So if you really want to stump somebody, ask him how many weights are needs to find the heavier ball between 6561 balls!!!
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