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May 26th, 2013, 11:51 AM  #1 
Senior Member Joined: May 2013 Posts: 109 Thanks: 0  A deck of cards
Someone breaks a 52 cards deck in two parts. Itīs known to be in the left part of the deck three times more red cards than black ones. Which is the MINIMUM number of red cards to be removed from the right part of the deck to get three times more black cards than red ones there?

May 26th, 2013, 08:59 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: A deck of cards
4

May 26th, 2013, 09:12 PM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,941 Thanks: 985  Re: A deck of cards
In order for blacks in right side to be 3 times reds, then the blacks must be multiples of 3; so: Code: LEFT RIGHT 6r 2b 20r 24b : remove 12r, then 8r 24b 15r 5b 11r 21b : remove 4r, then 7r 21b ******** 24r 8b 2r 18b : can't be done 
May 27th, 2013, 08:56 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: A deck of cards Hello, pedro malafaya! Quote:
Let = number of black cards in the left stack. Then we have: We will remove red cards from the right stack. Then we have: We want:[color=beige] .[/color] [color=beige]. . . . . . . [/color] [color=beige]. . . . . . . . . [/color] [color=beige]. . . . . . . . . . [/color]  

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