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August 20th, 2010, 10:37 PM   #1
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bài toán khó nh?t viêt nam n?m 2010

?? bài:có 23 ??ng ti?n, 1 ??ng gi? ( không bi?t n?ng hay nh? h?n so v?i các ??ng th?t c̣n l?i)
v?i 3 l?n cân th?ng b?ng ?i?n t? (có hi?n ???c rơ ?? l?ch, n?u bên ph?i n?ng h?n bên trái th́ cân ?i?n t? hi?n th? ?? l?ch s? ?? d??ng , c̣n bên ph?i nh? h?n bên trái th́ cân ?i?n t? hi?n th? ?? l?ch là s? âm) t́m 1 ??ng gi? .
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September 8th, 2010, 04:49 AM   #2
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or problem in vietnam in 2010

topic: has 23 coins, a fake money (not that heavy or lighter than the real coins remaining)

three times with electronic weight balance (which is clearly the deviation between the two sides, if the left-right worse then the current is positive on balance, slightly longer if the right than left on the current balance of is the number of audio, while two sides are equal, the current balance of 0 degrees. find a counterfeit coin.
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September 9th, 2010, 03:09 PM   #3
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Re: or problem in vietnam in 2010

Hello, giacat!

You posted this problem so many times . . . You are either: (1) desperate or (2) silly.
I also had to translate your English into our English.


Quote:
There are 23 indentical-looking coins.

One is a fake coin and is lighter than a real coin.

Using a balance-scale, determine the light coin in three weighings.

This problem can be solved with [color=blue]27[/color] coins.



[color=beige]. . [/color]





[color=beige]. . [/color]
[color=beige]. . [/color]
[color=beige]. . [/color]




[color=beige]. . [/color]



[color=beige]. . [/color]
[color=beige]. . [/color]
[color=beige]. . [/color]







[color=beige]. . [/color]
[color=beige]. . [/color]
[color=beige]. . [/color]

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September 9th, 2010, 03:45 PM   #4
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Re: or problem in vietnam in 2010

OOOOOOOOOOOOooohhhh... I thought those posts were just political propaganda!
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September 9th, 2010, 05:06 PM   #5
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Given 23 coins, 22 of equal weight and 1 fake underweight coin, one can find the fake coin as shown below.

(1)  Weigh 8 coins against 8 coins. If they balance, weigh 3 more coins against 3 more coins.
       If they balance, the last remaining coin is fake.

(2)  Using only the lighter 8 coins, weigh 2 coins against two coins.
       If they balance, weigh the remaining two coins against each other to determine which is fake.

(3)  Weigh the two coins from the lighter group against each other. The lighter one is fake.

[Oops! I hadn't seen the duplicate topics when replying. I've removed the duplicates.]
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