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SINGAPURCANAC
25th Oct 2008, 09:56
Twelve balls.
Among them one has different weight.
Using pan determine which one.
Only three(3) measurement is allowed.

BlueDiamond
25th Oct 2008, 10:18
I have a fair idea how this could done using balancing scales but I don't know what you mean by "using pan." Could you explain that please?

Also, what I have in mind wouldn't tell you if the ball with the different weight was lighter or heavier than the others, you could only identify that it was different. Does that matter?

TURIN
25th Oct 2008, 10:46
Yup, lost me at the first hurdle.

Tried to google "pan" but got a multitude of references.

Is it some form of standard deviation calculation?:confused:

25th Oct 2008, 10:53
Pan scales, perhaps? Think of the scales of justice. End of my input upon realising I can do it, but not in three!:sad: Wot Bluey sez, as well

1DC
25th Oct 2008, 11:01
Assuming pan scales are balance scales.
1.Put six on each side and discard the lightest side.
2.Put half of the remaining six on each side of the scales and discard the lightest side.
3. You are left with three. Take any two and put one on each side of the scale, if they balance the one you didn't use is the heaviest.If one side is heavier then that is the heaviest..

Takan Inchovit
25th Oct 2008, 11:02
Simple. Ignore the rules and weigh each one individually.

Fesch
25th Oct 2008, 11:04
1DC

well he didnt say that the one we are looking for is heavyer, he said it has a different weight

edit: i know this riddle, it took me about an hour or so to solve it, but I dont remember the sollution:\

1DC
25th Oct 2008, 11:11
Fesch.. Buggar glad my grandson isn't reading this, i keep telling him that the reason most people get their Maths wrong is cos they don't read the question properly......

Cron
25th Oct 2008, 11:21
Only need one measurement, put pan in garden underneath upstairs window, drop all balls simultaneously from upstairs window, heaviest will fall faster, so the first to hit is the heaviest.

Regards

Cron

PLovett
25th Oct 2008, 11:49
Split 12 balls into 3 lots of 4 and measure. One group will be heavier or lighter than the other 2 groups. Note the weight of the 2 groups that are the same.

Take the 4 balls of the "odd" group and split into 2 by 2 groups and weigh. One of the groups will be heavier or lighter than half of one of the discarded "normal" groups from the first weigh. Note the weight of the discarded group.

Weigh the 2 balls from the "odd" group. One of them will heavier or lighter than half the weight of the discarded group from the second weigh.

BlueDiamond
25th Oct 2008, 11:50
Pan scales, perhaps? Think of the scales of justice.
I thought of that, Baron but since they're just called scales not "pan" or "pan scales", I wasn't sure. Perhaps, as Turin says, it's a form of calculation we don't know about.

Anyway, assuming that scales may be what SINGAPURCANAC means, this is how I worked it out.

First you need to identify the balls by giving them numbers 1-12 and you split them into three groups. Group 1 consists of balls 1,2,3 and 4, group 2 has balls 5,6,7 and 8, and group 3 has balls 9-12.

Your first step is to balance group 1 against group 2. If they balance then group 3 has the odd-weight ball. If they do not balance, you can discard group 3.

If group 3 has the odd-weight ball, then you weigh any three balls from group 1 or 2 against 9,10 and 11 from group 3. (This is your second use of the scales.) If they balance, then ball 12 is the odd one out. If they do not balance then weigh any two balls from the heavier side (your third use of the scales). If they balance, the ball you left out is the heavier one, if they do not balance, the scales will show which of the two you are weighing is heavier.

That's as far as I've got so far ... still working on the rest.

sisemen
25th Oct 2008, 13:09
I took "pan" to mean lavatory pan. Here is my solution:

Put all the balls into the pan. Flush. If 11 balls remain then the one that is not there is lighter. If only one ball remains than that is obviously the heaviest.

Things might get a little complicated if one has taken some time to think about the matter and have also completed the crossword. :E

keel beam
25th Oct 2008, 13:27
Think of "Panning" for gold, no need to measure weight.

Though size is not mentioned, I assume pellet size?

Only 3 measurements allowed - how about measuring for Size, Weight and Volume for each ball?

SINGAPURCANAC
25th Oct 2008, 15:30
Pan scales, perhaps

Yes. thank you. I was in hurry this morning so I didn't take proper word from dictionary.
I appreciate all your answers but it is still far away from result.
Some of post were close but we need result. Exact one.
Someone has made good point. It is not stated that one of the balls are heavier/lighter than others.
Just different weight.
Still waiting.....

Cron
25th Oct 2008, 20:58
TwoOneFour, he would agree, no vacuum. Try dropping a snooker ball and a ping pong ball from ya bedroom window.

Regards

Cron

Hobo
25th Oct 2008, 21:08
As posted by ORAC in 2002:-

Label the balls 1, 2, 3,..., 10, 11, 12 so that you can distinguish between and identify them using these labels. Weigh 1, 2, 3, 4 against 5, 6, 7, 8

They balance, so 9, 10, 11, 12 contain the odd ball. Weigh 6, 7, 8 against 9, 10, 11.

They balance, therefore 12 is the odd ball and so weigh 12 against any other to discover whether it is heavy or light.

9, 10, 11 are heavy and so they contain an odd heavy ball. Weigh 9 against 10. If they balance, 11 is the odd heavy ball, otherwise the heavier of 9 and 10 is the odd ball.

If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light ball.

5, 6, 7, 8 are heavy and so either they contain an odd heavy ball or 1, 2, 3, 4 contain an odd light ball. Weigh 1, 2, 5 against 3, 6, 10.

They balance, so the odd ball is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy ball.

3, 6, 10 are heavy, so the odd ball can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light ball.

3, 6, 10 are light, so the odd ball is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.

If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

SINGAPURCANAC
26th Oct 2008, 08:58
@ Hobo,Blue Diamond,
It seems that result is ok.
Just to verify at relevant place.