Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC/ C8 N2 b# _. {- x' K/ J- J: e# D
# U: I* [: j0 |9 N! v, u+ iFirst weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.2 @" m8 Z; t7 {& v, C( [% F
, g$ g0 e* @0 ]& X- ]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.
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9 N! @: y) W" y& `( ^0 V2nd weight....AAAB ^ ACCC, there will be 3 scenarios:
7 n( w1 Q% X& a( t公仔箱論壇(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.
1 i, @5 L) f4 C+ ZTVBNOW 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。(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)os.tvboxnow.com7 d/ J: `# c2 R# Y1 q5 D- X1 H# H
(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. |