## 12 balls

You have 12 balls, and a weighing machine with no standard weighs. One of those 12 balls is either lighter or heavier than the others. Spot the ball in 3 chances with the weighing machine.

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 hint: 2x6, 2x3, 2x1+1 or 2x4+4, 2x2, 2x1
 Sorry, I forgot to mention that you must also be able to say whether the ball is heavy or light.

## 12 balls

This "puzzle" has been posted a zillion times here...

 Blog Entries: 1 Recognitions: Gold Member Maybe not a zillion times, but more than 5 times.

 Quote by Xalos You have 12 balls
I've had that dream before. Kind of a nightmare, actually. It was impossible to walk or sit down.

 and a weighing machine with no standard weighs. One of those 12 balls is either lighter or heavier than the others. Spot the ball in 3 chances with the weighing machine.
Oh! Okay.

6 vs. 5 + 1 lighter. Get the lighter set.

3 vs 2 + 1 lighter. Get the lighter set.

Weigh any 2 of them. If they are equal, the 3rd one is lighter.

 Quote by Poop-Loops I've had that dream before. Kind of a nightmare, actually. It was impossible to walk or sit down.
Thank you. I actually have spittle on my screen.

 Quote by Poop-Loops 6 vs. 5 + 1 lighter. Get the lighter set. . . .
It is not ok: the defective ball could be in the heavier set .

 Oh craps. I didn't notice that it could be lighter or heavier.
 The puzzle is new to me, so here's my answer: Weigh 4 against 4. If they even out, select three and weigh against 3 from the remainder. If they cancel out still, and since there's 1 ball left, one more weighing will give away the answer. If not, then the foreign ball is within those 3 balls selected from the remainder. At this step, it should be clear whether the foreign ball is lighter or heavier than the others. Now say the said balls are A B C, and other random balls now known to be normal are X X X. Weigh A X against B X. If they even out, then obviously it's C, if not, then either A or B. Now if 4 against 4 doesn't cancel out, substitute one side with the remainder. Depending on what the scale reads, determine whether the foreign ball is lighter or heavier. In any case, you'll end with a 2 against 2 weighing, which will easily give away the foreign ball by swapping two balls from each side, and substituting for 1 ball from the remainder on one side.

 Quote by Kittel Knight This "puzzle" has been posted a zillion times here...
I've told you a million times: don't hyperbolize!

 Quote by Werg22 The puzzle is new to me, so here's my answer: Now if 4 against 4 doesn't cancel out, substitute one side with the remainder. Depending on what the scale reads, determine whether the foreign ball is lighter or heavier. In any case, you'll end with a 2 against 2 weighing, which will easily give away the foreign ball by swapping two balls from each side, and substituting for 1 ball from the remainder on one side.
I am pretty positive the bolded part would not work.

 You're right, what I wrote is senseless. Impossible to determine a foreign ball out of 4 balls with only one weighing. I'll try rectifying the solution, tomorow

 Quote by DaveC426913 I've told you a million times: don't hyperbolize!
You are right, Dave!
Even my mother has already told me this a billion times...

 Ok here's the correction: 4 against 4 doesn't even out Say we have on the heavier side A B C D and on the lighter side S T U V, and X represent a ball from the remainder. Now weigh S X X X against A T U V. If they even, out, the foreign ball is among B C D, and is heavier. If S X X X > A T U V, then the foreign ball is among T U V and is lighter. If S X X X < A T U V, weigh A against X. If A = X, then S is the foreign ball, and is lighter. If A > X, then A is the foreign ball and is heavier. Solved