[Guest]

One of 14 tennis balls is a bit lighter or heavier (you do not know which) than the others.

All balls are of same colour except one with different colour which is not the defaulter.

How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?

You can only balance one set of balls against another, so no reference weights and no weight measurements.

[Guest]

What's a defaulter??

[Guest]

i suspect he means the odd colored ball is not the mis-wieght ball.

i personally don't see how it's possible, 4 is the minimum as far as i can see.

you basically have 13 balls then.

wiegh 4 balls against 4 balls.

--if equal, the odd ball is in the last 5 balls.

--wiegh 3 of the 5 agaisnt 3 of the normal balls.

----if unequal based on wiether it went up or down you know its heavy or light.

----wiegh 1 ball against 1 ball of the unequal 3.

------done.

----if equal then the heavy/light is in the remaining 2.

----wiegh 1 of the balls against a normal ball.

------done.

thats the easy part. now for the hard part.

--if unequal, then you have 8 balls any of which could be the light or heavy.

--wiegh 2 from each group agianst 4 normal balls.

----if unequal, then based on the result, you know its heavy or light.

----wiegh 1 ball against 1 ball of the 4. if equal then you have 2 left. one more wieghing required.

----if equal, then the 4 balls that remain must constian the heavy/light.

----wiegh 2 balls against two normals.

------if unequal then 1 more wieghing required.

------if equal wiegh 1 fo the remaining against a normal.

[plainglazed]

Label the balls 1-13 and X with X as the defaulter (not heavy or light but "normal").

Weigh 1,2,3,4 vs 5,6,7,8

if 1,2,3,4=5,6,7,8 then the odd ball must be either 9,10,11,12,13

weigh 9,10,11 vs three of the balls now known to be normal

if > or < then one of 9,10,11 must be > or < let's say >

weigh 9 vs 10 to see which is heavier (in this case) or if equal then 11 is heavy

if 9,10,11 = three of the balls now known to be normal then the odd ball is either 12 or 13

weigh 12 vs X to see if it is heavier or lighter or if equal, then 13 is the odd ball (cant tell if heavier or lighter though)

if 1,2,3,4 > 5,6,7,8 then the odd ball must be one of those

weigh 1,2,5 vs 3,4,6

if equal then the odd ball is light and either 7 or 8

weigh 7 vs X to see if it is the light ball or if 8 is light

if 1,2,5 > than 3,4,6 then either 1 or 2 is heavy or 6 is light

weigh 1 vs 2 to see if one is heavy otherwise 6 is light

similar strategy should work if 1,2,3,4 < 5,6,7,8 or 1,2,5 < 3,4,6

Edited December 17, 2010 by plainglazed

[benjer3]

Weigh 4 versus another 4, not using the differently colored ball.

If they are not balanced, weigh 2 on the heavier side versus the other 2, and then weigh the 2 on the heavier side against each other to find the heaviest.

If they are balanced, weigh 2 of the reamaining balls against the another 2, again excluding the differently colored ball. If they are again balanced, the ball that is the same color as the rest is the heaviest. If they are not balanced, weigh the 2 on the heavier side against each other to find the heaviest.

[benjer3]

Oh sorry, I didn't see it could be heavier OR lighter. I'll have to get back to you on that....

[Guest]

Yes the "not defaulter" means it is not a mis-weight ball..

[Guest]

Not Solved yet....

[plainglazed]

This strategy I think is better in that it defines the odd ball as heavy or light in all cases and requires the differently colored "defaulter" ball.

Again label the balls 1-13 and X with X as the defaulter (not heavy or light but "normal").

Weigh 1,2,3,4,5 vs 6,7,8,9,X

if 1,2,3,4,5=6,7,8,9,X then the odd ball must be either 10,11,12,13

weigh 10,11,12 vs three of the balls now known to be normal

if 10,11,12 > or < three known normal balls, one of 10,11,12 must be > or < let's say >

weigh 10 vs 11 to see which is heavier (in this case) or if equal then 12 is heavy

if 10,11,12 = three of the balls now known to be normal then the odd ball 13

weigh 13 vs X to see if it is heavier or lighter

if 1,2,3,4,5 > 6,7,8,9,X then the odd ball must be one of 1 - 9

weigh 1,2,6 vs 3,4,7

if 1,2,6 = 3,4,7 then either 5 is heavy or 8 or 9 is light

weigh 8 vs 9 to see which one might be lighter or, if equal, then 5 is heavy

if 1,2,6 > 3,4,7 then either 1 or 2 is heavy or 7 is light

weigh 1 vs 2 to see which one might be heavy or, if equal, then 7 is light

if 1,2,6 < 3,4,7 then either 3 or 4 is heavy or 6 is light

weigh 3 vs 4 to see which one might be heavy or, if equal, then 6 is light

a similar strategy should work if 1,2,3,4,5 < 6,7,8,9,X

[araver]

Plainglazed, your strategy works .

I had not believed for a moment that the last case can be dismissed due to symmetry as it looks asymmetrical: one of the sides has the defaulter on it. But it does work out with those exact second and third weighings and mirror heavy/light arguments.

[Guest]

Good work Plainglazed