28 replies to this topic

[rookie1ja]

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.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
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Spoiler for Solution

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.

Spoiler for old wording

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.

[nzhunter]

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???

[netwebdave]

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.

[nzhunter]

Got it now. Cheers.

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

[psyber859]

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.

[akshay.nta]

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

[Rudy22]

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

[bigbic]

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.

[TorqueRat]

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.

[JASON4P8C10]

Spoiler for ...

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
DISCARD THE LIGHTER 3
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

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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