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May 26th, 2013, 10:51 AM   #1
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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?
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May 26th, 2013, 07:59 PM   #2
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Re: A deck of cards

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May 26th, 2013, 08:12 PM   #3
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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
********So removing 4 reds is minimum
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May 27th, 2013, 07:56 AM   #4
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Re: A deck of cards

Hello, pedro malafaya!

Quote:
A 52-card deck in broken into two stacks.
The left stack has three times as many red cards than black cards.
What is the MINIMUM number of red cards to be removed from the right stack
to get three times as many black cards than red cards there?

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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