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# Weighing IV.

35 replies to this topic

### #21 Eru

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Posted 22 February 2008 - 11:17 PM

My graphical representation of the solution for only 3 weighs... might be a little difficult to read, but see if you can follow.

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### #22 sebenak

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Posted 25 February 2008 - 09:28 AM

3 is the minimum number of uses of the scale. My basketball coach in high school told the team that if anyone could solve this puzzle the team would avoid running excessive sprints that week. His version was a tad different, (he wanted the method, and told us we could only use the scale 3 times) but the principles are the same. After solving for three weighings I don't believe it's possible to get enough information with only 2 uses of the scale.

The key to solving this (as "bonanova" shows) is realizing that you get more information out of the unbalanced scale than most people realize at first. Say you put to piles of four on either side of the scale. If it is uneven, you know that the odd ball is one of those eight (obvious). What wasn't as obvious (to me) at first was that you also know, that if the odd coin is later determined to be from the pile which is now on the low side of the scale, IT IS LIGHTER than the rest. Vice versa for the high side.

For whatever reason having to mix the "lighter than", "heavier than", and "known normal" balls was a no brainer once i realized this. Otherwise my solution was the same as Bonanova's noted above.

I really like the array solution though, brilliant Stephcorbin thanks!
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### #23 luckywassabi

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Posted 12 March 2008 - 10:33 PM

I miss the "what get's wetter as it dries" riddles, i don't even attempt these ones. But still, nice job of accmplishing i didn't even bother to !
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### #24 el_duro_jr

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Posted 25 April 2008 - 07:05 PM

DEVIDE INTO TWO EQUAL GROUPS......6 & 6........REPEAT STEP FOR LIGHTER SIDE....3&3........THEN TAKE THE SIDE THAT IS LIGHTER AND TAKE 2 ON ONE SIDE AND ADD1 TO THE 1 U HAVE REMAINING .AND DEVIDE THE SIDE THAT IS LIGHTER TO FIGURE OUT WHICH BALL
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### #25 Smile of God

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Posted 20 May 2008 - 01:10 PM

The key to the solution is "the least " amount of times you must weigh to determine which ball is heavier or lighter. So it is possible on the first comparison of any two balls you find imbalance. That is the first weighing. By removing one of the two and replacing it with another, by process of elimination you discover the odd ball. If you have equilibrium, it is the first ball you removed. If you have the same imbalance, it is the remaining ball that was not first removed. It is not possible to arrive at the answer with less than two weighings.
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### #26 shobith

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Posted 04 June 2008 - 05:56 AM

we can find it in atmost 3 weighings.(solution is already given by moderator)
There is a formula to solve this kind of problems( i got this formula from combinatorics, and found suitable for these kind of problems)
no. of uncertainties =x
no. of outcomes for a weigh =y
no. of weighings required =n
then y^n >x ( the minimum value of n satisfying the equation is the answer)
(this formula would be better for competitive exams)
for this problem;
no. of uncertainties = 2*12=24 ( each ball is heavier or lighter, we dont know)
no. of possible outcomes of each weigh=3 ( each weigh says whether one half is heavy,light or equal to other)
3^n>24 ( 3 is the min. value of n satisfying the equation)

Edited by shobith, 04 June 2008 - 05:57 AM.

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### #27 reelfishy

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Posted 07 September 2008 - 02:14 AM

To clarify:

The solution of weighing 6 and 6, 3 and 3, and then finding the oddball doesn't work. When you get one group to be heavier and one group to be lighter, you still don't know whether the oddball is heavier or lighter-- if it's heavier, you want to start weighing the heavy group, if it's lighter, you want to start weighing the lighter 6.

start with 5 balls on either side and go from there. 3 tries will do it.
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### #28 Louie

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Posted 12 January 2010 - 03:11 PM

Could you do this same riddle with more balls, or is 12 the max? I think it is possible to determine the odd ball of 13 with still just 3 weighings. The problem would be determining if it was lighter or heavier...

What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24 because if you weighed 6 v 6, worst case you would have 12 unknowns with 3 weighings left (same as this problem). Or you would have 6 possible heavys and 6 possible lights which is infomation you could use. So you could probably do 26 or something.
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### #29 cho3eiV8

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Posted 24 March 2010 - 07:36 PM

Could you do this same riddle with more balls, or is 12 the max?

Actually, it is possible to have up to 14 balls and determine which one is lighter/heavier in only three weighings. This makes the problem even more interesting; go ahead and try it
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### #30 cho3eiV8

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Posted 24 March 2010 - 08:07 PM

What would be the max number of balls you could have if you had 4 weighings? My first thought is more than 24...

The maximum number of balls for which you can solve the problem in 4 weighings is in fact 39.

If you have an additional ball that you know is "normal", you can do up to 41 balls. My previous reply assumed such an "extra" ball. Without the extra ball, the maximum number of balls for which you can solve the problem in three weighings is actually 13, not 14.
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