My Math Forum bài toán khó nh?t viêt nam n?m 2010

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 August 20th, 2010, 10:37 PM #1 Newbie   Joined: Aug 2010 Posts: 2 Thanks: 0 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? .
 September 8th, 2010, 04:49 AM #2 Newbie   Joined: Aug 2010 Posts: 2 Thanks: 0 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.
September 9th, 2010, 03:09 PM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: or problem in vietnam in 2010

Hello, giacat!

You posted this problem so many times . . . You are either: (1) desperate or (2) silly.

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.

$\text{Divide the coins into three equal subsets:}$
[color=beige]. . [/color]$\underbrace{\{\circ\:\circ\:\circ\:\circ\:\circ\:\ circ\:\circ\:\circ\:\circ\}}_A \;\;\underbrace{\{\circ\:\circ\:\circ\:\circ\:\cir c\:\circ\:\circ\:\circ\:\circ\}}_B \;\;\underbrace{\{\circ\:\circ\:\circ\:\circ\:\cir c\:\circ\:\circ\:\circ\:\circ\}}_C$

$\text{Weighing 1: weigh set }A\text{ against set }B.$

[color=beige]. . [/color]$\text{If }A\,<\,B\text{, take set }A.$
[color=beige]. . [/color]$\text{If }A\,>\,B\text{, take set }B.$
[color=beige]. . [/color]$\text{If }A\,=\,B\text{, take set }C.$

$\text{Divide the 9 coins into three equal subsets.}$
[color=beige]. . [/color]$\underbrace{\{\circ\:\circ\;\circ\}}_D\;\;\underbr ace{\{\circ\:\circ\:\circ\}}_E\;\;\underbrace{\{\c irc\:\circ\:\circ\}}_F$

$\text{Weighing 2: weigh set }D\text{ aginst set }E.$

[color=beige]. . [/color]$\text{If }D\,<\,E\text{, take set }D.$
[color=beige]. . [/color]$\text{If }D\,>\,E\text{, take set }E.$
[color=beige]. . [/color]$\text{If }D \,=\,E\text{, take set }F.$

$\text{We have three coins: }\;G,\;H,\;I$

$\text{Weighing 3: weigh }G\text{ against }H.$

[color=beige]. . [/color]$\text{If }G\,<\,H\text{, then }G\text{ is the light coin.}$
[color=beige]. . [/color]$\text{If }G\,>\,H\text{, then }H\text{ is the light coin.}$
[color=beige]. . [/color]$\text{If }G \,=\,H\text{, then }I\text{ is the light coin.}$

 September 9th, 2010, 03:45 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: or problem in vietnam in 2010 OOOOOOOOOOOOooohhhh... I thought those posts were just political propaganda!
 September 9th, 2010, 05:06 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,629 Thanks: 2077 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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