A fine example of rationality getting better and better, as more people contribute.

You really have to read this whole thing through to fully enjoy the final contribution.

Basically, cantfindongoogle.com is a site that people go to when they... can't find stuff on google. And people usually reply, and help out.

Chekkit.....

joshua at 01/18/2006 03:59:51 pm

Really Looking For: The answer to a math question!

- There are 8 billiards balls. They are all identical. Same color and everything. Exactly the same as each other. Apart from 1 ball that is different in weight. Weather lighter or heavier, we don't know.

- You have a scale to weigh items. You can only use the scale 3 times.

- You may weigh any combination of balls on either side. (8/1, 6/2, 4/4, etc etc.)

- You also do not need to put all balls on the scale. (For example, you canhave 3 balls on each side of the scale, and two off the scale)

Search Terms Tried: 8 ball algebra question

Remarks: I've been wracking my brain on this since last friday and can't figure it out. It's for a class, no extra credit and it' snot a grade. just something my profesor came up to us with and it's driving me insane!

Category: Math (college algebra)

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==COMMENT 01==

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follow the steps: first time: 4 on each side - 1 side will be heavier, it MUST be because only 1 ball is heavier second time: take the four on the HEAVIER side off, put 2 on one side, two on the other, one set of two will be heavier. third time: take the two from the heavier side, put one on each side, one will be heavier.

have a good one.

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==COMMENT 02==

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Here's how to do it only using the scale twice (see if your professor can figure it out).

Put 2 of the balls aside. Split the 6 balls up evenly and weigh them (first use of scale). If they are equal, one of the 2 balls that you set aside is the one. Weigh them to find out which of them is heavier (second use of scale). OR...If one of the 3 ball sets was heavier, remove one of the balls from the heavier set. Weigh the remaining two balls from the set to find out which one is heavier (second use of scale). If they are equal then you know it is the one ball that you put aside.

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==COMMENT 03==

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Neither of the above processes would work because they are both assuming that the odd ball is heavier. The problem states that we only know one ball is a different weight. We have to identify it, yes, but we must also determine whether it is lighter or heavier.

Using W's method, the scale could balance on the 2nd use, meaning that one of the other 4 was the odd ball and it was lighter. That leaves you one more use to determine which of the 4 it was. It won't work.

Using Matt's method, when you weigh 2 of the 3 balls from the heavier set (second use) and they balance, it does not necessarily tell you that the 3rd ball is heavier, because the entire set may be the same weight, and it only tilted the scale because a lighter ball was on the other side.

The real solution is slightly more complicated.

To make things easier, I will number the balls (1-8).

Place balls 1 & 2 on one side of the scale, 3 & 4 on the other, and 5,6,7 & 8 off to the side. This is your first use of the scale. This gives you one of two scenarios. In scenario A, the scale slants, meaning the heavier or lighter ball is among that group (1,2,3, or 4). In scenario B, the scale balances, and we know that the odd ball is in the other group (5,6,7, or 8).

Following scenario A, knowing the odd ball is either 1,2,3, or 4, we place 3 of the 4 balls on one side of the scale (let’s say 1,2, & 3) and 3 from the other group (5,6,7, 8) on the other side. Since we already know 5,6,7, & 8 are not the odd balls, if the scale is even, then the 3 balls used must also be of uniform weight. Therefore we now know ball 4 is the odd ball. We need now only measure ball 4 against any other ball (for our third use of the scale) to see if 4 is lighter or heavier. If, however, the scale slants on the second try, not only can we eliminate one more ball (ball 4) as the odd ball, but we now know whether the odd ball is heavier or lighter because we know which side of the scale contains the 3 uniform balls. In this scenario we would then weigh ball 1 against ball 2. If it balances, ball 3 is the odd one; if it slants, we use our knowledge gained from the 2nd use of the scale about whether the odd ball is heavier or lighter to determine which is the odd ball based on which way the scale slants.

In scenario B, the scale balances on the first use, so we determine that one of the group 5,6,7,8 is the odd ball. We then go through the same process as above, only swapping the roles of the two groups. Let me know if this is not clear enough, but I assure you it works.

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==COMMENT 04==

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I did think about that and certainly didn't have the answer you provided (well done). Given time maybe I would, but not entirely certain. Therefore, I went with the assumption that the poster's comment was verbose in that..."...to use the balls in different variations to figure out which one of the eigth is different."

My solution provides the answer to which one is different but certainly not if it's heavier or lighter.

Again, well done.

Added later: thinking about this, my solution wouldn't work for different, just if heavier is known. I stand corrected.

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==COMMENT 05==

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First measure the first three and the last 3:

If it is equal you have an easy job: measure the two which left - the heavier is the odd one out. If not equal choose two from the heavier side of the first measuring and put on the scale( A side). At the other side put the last ball from the heavier side of the first measuring and one from the two unmeasured balls to the B side of the scale. Now if B side is heavier, we have the odd one out - tha ball we put on B side first. If equals - the odd one out is not heavier but lighter tha the others. In this case I do not know what to do But someone cloud go along this logics.:) If A side is heavier - we have a last measuring attempt: compare the two balls of the A side - the heavier is the odd-one-out.

That is all.

Regards, Viktor L. Takacs

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==COMMENT 06==

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This method would work as well. You just stopped shy of a complete solution. You are not sure what to do if the 4 balls measured on the second use of the scale balance, but it can be resolved with one use because we would now know the odd ball is lighter and in the first group of three balls we weighed. So, for the last use of the scale, we measure two of those first three balls. If the scale tips, the lighter one is the odd ball; if it balances, the one of the three that is not on the scale has to be the odd ball.

Good work.

One other thing: you say that if the first 6 balls measured are equal then you just weight the last 2 against each other and the heavier one is the odd one. This is incorrect, but it still doesn't kill your method. If we weigh them against each other, we don't know which is odd because we don't know whether the odd is heavier or lighter yet, but, in this scenario, we still have two uses of the scale left. One need only measure each of the two balls against one of the 6 uniform balls to determine which is odd and whether it is lighter or heavier.

The interesting thing about this method is that it allows for a couple of ways in which you could find the answer in two uses with a little luck.

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==COMMENT 07==

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Pick up each of the balls. The one that is heavier or lighter than the rest is heavier or lighter than the rest.

I'm cutting the Gordian knot a bit, but it works.

=====Transmission ends=====

Pure classic

:D

## Friday, June 15, 2007

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