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Bally Heck
10th Dec 2001, 00:49
You are given a set of balance scales and nine aparently identical billiard balls. Only one of the balls is a different weight (JAR-OPS=Mass).

You may use the scales but three times balancing ball(s) against ball(s) in order to determine which is the errant ball and whether it is lighter or heavier.

Gash Handlin
10th Dec 2001, 01:30
You put three balls in each side of the scales.

If the scales balance then the light ball is one of the remaining three. Otherwise it is obviously in the side that rises in the scales!

then you place 1 of the set of three balls which includes the lightest in each side of the scales and again if they balance then the lightest ball is the one left over otherwise its the one in the scales that rises.

(only used the scales twice ;) )

henry crun
10th Dec 2001, 01:37
Gash Handling, the question is only answered when it covers the odd ball being lighter or heavier.
Try again. :)

Gash Handlin
10th Dec 2001, 01:43
oops, your correct Mr Crun, but the principle is still the same, work on three balls at a time, eliminate the two sets that cant possibly contain the odd ball and then repeat for the set of balls which does contain the ball

Bally Heck
10th Dec 2001, 15:17

Eliason
10th Dec 2001, 16:24
OK...
9 Balls - split them up: 3-3-3
put 2 sets on the scale - either both are the same weight - or one is heavier/lighter.
Do the same again - just exchange 1 set of the balls.
If before they were the same weight - now you should know which set is the one with the wrong ball (the 3rd) - and whether its heavier or lighter (goes up or down).

If before you had one set which was wrong - you should discover now the same - cause the one you compare it too will be all standard.

Ok. Now you have 3 balls remaining - and know, that one of those is heavier/lighter.

Just place any 2 of them on 1 side of the scale. Now you know which it is :)

Grainger
10th Dec 2001, 16:29
OK Bally - Gash has it OK, but to spell it out:

Divide the balls into three sets of three. Call them A, B and C

Weighing 1: A - B
Weighing 2: A - C

Now either you'll find that both weighings are unbalanced, in which case A contains the odd ball, or else A=B and thus C contains the odd ball, or A=C and B contains the odd ball.

By this time you'll also know whether the odd ball is lighter or heavier.

Take the set with the odd ball and label the balls x, y and z

Weighing 3: x - y

if x = y then z is the odd one out
if x &lt;&gt; y well you know by now whether the odd one is lighter or heavier so we're done.

Comprehensive enough ?

gravity victim
10th Dec 2001, 16:36
This is not the first thread on PPRune that could be described as a lot of balls, but it's the first completely sensible lot of balls! Now I have the solution, I'm off to try it on the kids - thanks.

I Am Ugly
10th Dec 2001, 16:36
I remember this giving me a headache a few years ago.

But the problem I was posed was 12 balls and 3 weighings. Try it.

I still vaugely remember the solution, but it's about as hard to explain as it is to work out.

Bally Heck
10th Dec 2001, 20:43
10 out of 10 Eli and Grainger. Gash is assuming that one is lighter which aint necessarily so. Of to try the twelve ball scenario. :confused:

Grainger
10th Dec 2001, 21:41
Hm

you could divide the 12 into 3 lots of 4 and then follow the same process - but you'd end up with two pairs and thus possibly four weighings :confused:

sanjosebaz
11th Dec 2001, 11:32
The twelve ball problem is tortuous! The solution can be found in this SPOILER (http://newton.dep.anl.gov/newton/askasci/1995/math/MATH007.HTM)