nightrider

Not very hard. Did it in 2 hrs. Now let me verify.

nightrider

put 4 & 4 is a good start. Your answer is right when the problem ball exists in team A or team B. But when the problem ball exists in team C then your answer is wrong because you end up not knowing whether the problem is heavier or lighter so you still don't know which one is the problem ball.
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: Y( J# J3 @0 Z% W$ }; D5.39.217.77:8898The correct answer is after the first weight, (4 balls on each side and balance), you know the problem ball is among the remaining 4 balls (team C). Take 3 balls from team C, weight against 3 normal balls. If balance, then the problem is with the remaining 4th team C ball, just weight against any normal ball, then you know it's heavier or lighter. If not balance then note heavier or lighter. Just take 2 from the 3 and weight against each other, if balance it's the third ball and you already know it's heavier or lighter. If not balance then the problem ball is either the heavier ball or the lighter ball base on the previous weight.

nightrider

Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC
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8 P& t- i5 B9 _. x3 M# W  P& H公仔箱論壇First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.TVBNOW 含有熱門話題，最新最快電視，軟體，遊戲，電影，動漫及日常生活及興趣交流等資訊。0 p; X9 M5 m% G+ z0 g* E& u  r
tvb now,tvbnow,bttvb1 M, T# Z2 p3 m: U& [- V
From the first weight, let's say AAAA is heavier then BBBB, then either the problem ball is heavier among AAAA, or lighter among BBBB, but all C balls are normal.公仔箱論壇2 n* W+ N5 |8 q! Q) Z! b9 X
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2nd weight....AAAB ^ ACCC, there will be 3 scenarios:tvb now,tvbnow,bttvb9 H, t" [. p6 x9 A0 ^3 N* R
(1) if AAAB is heavier then ACCC, for sure the problem ball is a heavier ball and it's among AAA. Take 2 of these and weight against each other, the heavier ball is the problem, if balance then it's the remaining ball.TVBNOW 含有熱門話題，最新最快電視，軟體，遊戲，電影，動漫及日常生活及興趣交流等資訊。6 M2 p& @& Y: t  S/ y: I; W* l
(2) if ACCC is heavier then AAAB then weight the A ball in ACCC against a normal C ball, if balance then the problem ball is the B ball in AAAB and it's a lighter ball. If heavier then this is the problem ball. (note it cannot be lighter)
$ e& D2 M7 r, @8 J5 V: |6 {" W(3) if balance then the problem ball is one of the 3 B balls not touched in the 2nd weight and it's a lighter ball. Take 2 of these B balls and weight against each other. If balance then the other B ball is the problem. If not balance then the lighter ball is the problem.

nightrider

I wanna tell you I use my brain.

nightrider

Yeah Yeah! LMAO!

nightrider

Most answers missed the point that the problem ball could be lighter or heavier then the rest.