Okay let's try again, first divide the 12 balls into 3 groups.....let's name them AAAA, BBBB & CCCC- L2 ^2 x- G2 i6 A+ w! k+ L
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First weight....AAAA ^ BBBB, if balance, then problem in C balls, see my post above. If not balance, continue below.3 V: h4 S$ y8 } u8 `2 l+ `* h
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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8 ]* `/ a: m) W1 m( i; a* ?
# W3 i: v7 j* h1 R6 T; O$ g2nd weight....AAAB ^ ACCC, there will be 3 scenarios: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。: E6 [: j' A$ d' }2 s8 ~5 K
(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.
8 M+ Z& W+ m: s* ~(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)
- C5 z2 \+ c: g$ W) S, r公仔箱論壇(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. |