月※月

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[ 本帖最後由 月※月 於 2007-11-8 04:52 PM 編輯 ]

vlun

TRY TRY

razorking123

猜到了。。先看看答案:onion05: :onion05:

adeline_miao

thxxxxxxxxxxxxxxxx

l__eo

thx for sharing

jamesdean

i thing the answer is( M* \: O, v# v, l) C4 `* t, V
start with 4 on each side of the scale,this will have two possible outcome, same or diff.3 B$ v1 m  k2 }  C# `
If the same then 4 that left over contain the odd ball. put 1 of the 4 to each side, two outcome, same or different again. Replace 1 from the remaining, if the same then left over is the odd or if different the replacement is the odd one. I didn't time myself but I think I take less than 30 minutes. There are two more possible out come but still need three weighing but I think I am running out of space.

jamesdean

stage 3c and 3d is not possible since all balls look identical, there is no way logically to define which one is chosen from normal batch. So answer do not stand.

jamesdean

if you can identify which balls come from normal batch as stage 3c mentioned, then a much easier method can be used. If first 4 balls weigh different, then remove 2 on each side, take one normal batch C to each side. then choose either zone balls that remove earlier and place back one on either side. If the same then what ever left untouched contain the odd ball. weight against one ball from normal batch C to find out. If different, then odd ball in scale. Two possibility, balls that reintroduced back to the scale contain the odd ball or remaining balls from the beginning containing the odd balls. if reintroduced balls contain the odd one then weigh against C. Like stage 3C you should be able to tell the different between identical balls. Well I don't see the point of continuing as There is a flaw with this question. So good luck identify the identical balls.

win333

:onion10: thanks

noem123

let me see

milk1111

Rinoas

lets try it out :P

zzylovejay

30 以内
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AnT_727

seeeeeeeeeeeeeeee