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# Weighing VII.

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Weighing VII. - Back to the Water and Weighing Puzzles

Given 27 table tennis balls, one is heavier than the others.

What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? Always. Of course, the other 26 balls weigh the same.

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Weighing VII. - solution

It is enough to use a pair of scales 3 times.

Divide the 27 balls to 3 groups, 9 balls in each. Compare 2 groups – the heavier one contains the ball. If there is equilibrium, then the ball is in the third group. Thus we know the 9 suspicious balls.

Divide the 9 balls to 3 groups of 3. Compare 2 groups, and as mentioned above, identify the group of 3 suspicious balls.

Compare 2 balls (of the 3 possibly heavier ones) and you know everything.

So we used a pair of scales 3 times to identify the heavier ball.

Having 27 table tennis balls, one is heavier than the others. How many times (minimum) do you need to use a pair of scales to identify it.

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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• 1 month later...

Hey there

You've got me thinking...

So we came up with 2 different ways to do this, using the scale (we think) only 2x but also 3x too. However, I'm a wee bit confused having read your answer...

Your method states using the scales 3x but in each of the 3 identical steps you state, "Compare 2 groups"... how do you compare 2 groups by using the scales only once?

Logic >>

Step 1: 3 sets of 9 balls, and

Step 2: 3 sets of 3 balls, and

Step 3: 3 sets, 1 ball in each.

Weigh 1 set, you have one weight but you still do not know if you have the heavier ball or if it is with the remaininig 2 sets or 18 balls = you must weigh another of the sets (either 9 balls or the entire 18 balls) to determine which set contains the heavier ball (i.e. you either get an equilibrium or not) >> You used the scale 2x.

Am I missing something here??? It would appear, if I am not incorrect, your method uses the scales 6x???

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This is talking about the scale that is used to compare two groups. I know I will be banned for using this example but if you ever have seen Celebrity Fit Club, one team stands on one side, the other team on the other and it compares the weight of the two. The heavier group then lowers. It's about the same as a balanced teetor totter with two children. If they both do not move, the heavier side will go to the ground.

So, if you have 3 sets of items, you can weigh two of them. If they are center balanced, then you know the third set is heavier because in this case, one item has to be heavier than the others. Hence, if you had 3 sets of 9 and applied what was stated, you will eliminate 18 objects leaving 9 in one use of the scale. Apply the same theory 2 more times and you are down to 1 object. Hope this helps.

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Got it now. Cheers.

Will work on our method of using an e-scale only twice!

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• 5 weeks later...

Actually, if you're very lucky, you might get away with one weighing.

Split the balls into 2 groups of 13, with one ball left out. The heavier group on the scale contains the heavy ball. If both are equal, the odd ball not weighed is the heavy ball.

If you're not lucky, the 13 remaining balls can be broken into different (numeric) groups that still allow you to finish in three steps.

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• 4 weeks later...

27 = 9 + 9 + 9

step 1 divide in three equal part and weight two part if both are equal then third one has default ball

if we get default ball part and again divide in three equal part

9 = 3 + 3 + 3

step 2 again do same process step 1

3 = 1 + 1 + 1

step 3 follow the step 1

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Split the balls into 2 groups of 13, leaving 1 ball unmeasured...if your lucky, the scales will be exactly equal leaving the left out ball the heavier one...so maybe only 1x

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nzhunter , your thinking of a normal weight scale, where you put an object(s) on and it gives you the weight of it, the riddle is talking about a balance scale, which has two sides to put weight on to compare them.

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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.

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• 1 month later...

WEIGH 13 AND 13 IF THEY ARE EVEN THEN THE 1 LEFT IS HEAVY

IF NOT TAKE THE 13 LIGHTER AND 1 EXTRA AND DISCARD

TAKE 6 OF THE ONES LEFT AND PLACE ON OTHER SIDE OF SCALE

REMOVE THE 7TH FORM THE 1ST GROUP

ONE AGAIN IF THEY ARE EVEN THE 1 LEFT IS HEAVY

IF NOT TAKE DISCARD LIGHTER GROUP AND EXTRA

3 FROM THE 1ST GROUP PUT ON OTHER SIDE

TAKE 1 FROM THE GROUP AND PUT ON OTHER SIDE OF SCALE, TAKE 1 AND HOLD

EITHER 1 ON THE SCALE IS HEAVY OR ONE IN YOUR HAND

SO THE MINIMUM NUMBER OF WEIGHS ASSUMING YOU DO NOT GET LUCKY AND AND END UP WITH THE HEAVY BALL AS AN EXTRA IS 4

THANK YOU! THANK YOU!

I TAKE DONATIONS

i didnt feel like retyping everything the above is for 1 scale but for "a pair of scales" assuming it isn't like a pair of pants and acutally means 2 scales the minimum weighs is 3

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

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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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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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• 4 weeks later...

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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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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• 2 months later...

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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• 1 month later...

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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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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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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• 4 weeks later...

You have to do it 26 times otherwise you don't know which one is the heavy.

Example-

Say i said the answer was 1 time...9 groups which i weigh. Then if one is heavier then that is the group.

If there is equlibrum then the group unweighted is the heavier one.

BUT-this means i only know the group, not the specific ball.

So I have to weigh 26 times.

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• 2 weeks later...

You could actually accomplish this in just one weighing using the simplest scale possible. Imagine a teeter-totter perfectly centered on its fulcrum(simple balance scale)....you align the balls however you like on the scale . plugging a few numbers into a physics equation will tell you exactly which ball is the heaviest.Thinking Equation would look something like this B1 D1+B2 D2+B3 D3 +B4 D4= B5 D5+ B6 D6....etc...etc... Where B is the weight(or ball) and D is the distance it is from the fulcrum.

I could be wrong....sounded good to me though ::shrugs::

Edit: I'm really stoned, that wouldn't work...maybe a more advanced equation adding variable/constant for gravity, and the amount the scale traveled. Perhaps someone out there is experienced at physics and gimme advice on this method.

Edited by hendle
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• 2 weeks later...

You know the scale has three outcomes, L = left heavy, B = balanced, R = right heavy

Given that we know the odd ball is heavier than the rest, with a good choice of n comparisons, you can guarantee to identify the heavier ball out of 3^n balls. (3^n means three raised to the power of n, so 3^3 = 3 x 3 x 3 = 27 meaning only three weighings.

That's actually all you need to answer the question of how many trials to always be sure of the correct answer.

Rather than use the result of each weighing to choose which balls to weigh next, you can decide to always weigh one third of the total number of balls on each side of the scales and construct a truth table with all 27 unique combinations of three weighing trial results, meaning you can predetermine the sequence of trials you'll run rather than having to wait for the result of the first trial to determine the balls you weigh on each side at the second trial etc.

Here's an example truth table, working systematically through all 27 permutations and assigning a meaning, which determines whether the ball in question goes on the left side, the right side or is omitted for each trial.

Results in order:

RRR = ball 1 is heaviest

RRB = ball 2

RRL = ball 3

RBR = 4

RBB = 5

RBL = 6

RLR = 7

RLB = 8

RLL = 9

BRR = 10

BRB = 11

BRL = 12

BBR = 13

BBB = 14

BBL = 15

BLR = 16

BLB = 17

BLL = 18

LRR = 19

LRB = 20

LRL = 21

LBR = 22

LBB = 23

LBL = 24

LLR = 25

LLB = 26

LLL = 27

Other allocations of numbers to each outcome are possible of course.

Using this truth table, you can set up the three trials by looking down the list for L and R in the first column for the first trial, in the second column for the second trial, and the third for the third trial, which will have 9 balls (assuming balls are numbered 1 to 27) on each side and produce those results as follows:

```[font="Lucida Console"]
ON LEFT SIDE OF SCALES          ON RIGHT SIDE OF SCALES
FIRST TRIAL   19 20 21 22 23 24 25 26 27       1  2  3  4  5  6  7  8  9
SECOND TRIAL   7  8  9 16 17 18 25 26 27       1  2  3 10 11 12 19 20 21
THIRD TRIAL    3  6  9 12 15 18 21 24 27       1  4  7 10 13 16 19 22 25
[/font]
[/codebox]```

If you had come up with a different truth table you'd have a different set of trials of 9 balls. The ball with result BBB never gets put on the scales. In my truth table, that's ball 14.

Pick a number from 1 to 27 and see what the results would be in sequence, then look that up in the truth table. E.g. ball 23 will give the results Left-side heavy, Balanced, Balanced. The sequence LBB in the truth table is next to the number 23.

You can also apply the same technique to the 12 balls, only one of which differs but you don't know if it's heavier or lighter than the normal balls. You then have 24 truth table items ('ball is light' and 'ball is heavy' outcomes for each of the 12 balls, which must be unique). Certain truth table results out of the 27 possible won't give you the full answer (e.g. BBB is useless because it doesn't tell you if the ball is heavier or lighter, but you only need 24 outcomes). You then need to ensure that you assign the numbers to ensure you have 4 balls on each side for each trial. You also know that if ball 1 being heavy is LRB then ball 1 being light must have result RLB (i.e. L and R swap over, but B stays the same because that ball wasn't weighed).

You could envisage balls with a 1 gram difference on scales with a 0.5 gram blob of glue stuck to the right-hand side of the scales and unable to be removed or counterbalanced. This eliminates the possibility of balancing evenly leaving only 2 possible outcomes of each trial weighing (L or R) and makes the problem into a 2^n problem, e.g. 4 trials then allows you to identify which one of 16 balls is heavier than the rest. This is a binary truth table (e.g. 1 = L, 0 = R) more familiar to computer programmers.

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I think if you split it in half 13/13 and 1 out. if scales equal its odd one out (so minimum weighs)... (First weigh)

next - take heavier side and repeat 6/6/ 1out.... (Second time)

next take heavier side and repeat 3/3 (third time)

next take heavier side and you'll have 1/1/1out... so if the scale is equal - its odd one out, otherwise its the heavier of the two..

SO - Four Times Max needed.

******

OK - darn... its three by splitting into 3 groups versus two you get there one time faster. Great quick puzzle.

Edited by RiddleRookie
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0. Set fire to all balls and the one that finishes burning last is the heavier one

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This topic is now closed to further replies.