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 May 26th, 2013, 10:51 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, 07:59 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,962 Thanks: 1147 Math Focus: Elementary mathematics and beyond Re: A deck of cards 4
 May 26th, 2013, 08:12 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 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
May 27th, 2013, 07:56 AM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

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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 $b$ = number of black cards in the left stack.

Then we have:

$\;\;\;\begin{array}{c|c|c|}
& \text{Left} & \text{Right} \\ \\ & \text{stack} & \text{stack} \\ \\ \\ \hline \\ \\ \\
\text{Red} & 3b & 26\,-\,3b \\ \\ \\ \hline \\ \\ \\
\text{Black} & b & 26\,-\,b \\ \\ \\ \hline \end{array}$

We will remove $x$ red cards from the right stack.
Then we have:

$\;\;\;\begin{array}{c|c|} & \text{Right} \\ \\ & \text{stack} \\ \\ \\ \hline \\ \\ \\
\text{Red} & 26\,-\,3b\,-\,x \\ \\ \\ \text{Black} & 26\,-\,b \\ \\ \\ \hline \end{array}$

We want:[color=beige] .[/color]$\text{Black} \:=\:3\,\cdot\,\text{Red}$

[color=beige]. . . . . . . [/color]$26\,-\,b \:=\:3(26\,-\,3b\,-\,x) \;\;\;\Rightarrow\;\;\; 26\,-\,b \:=\:78\,-\,9b\,-\,3x$

[color=beige]. . . . . . . . . [/color]$3x \:=\:52\,-\,8b \;\;\;\Rightarrow\;\;\;x \:=\:\frac{52\,-\,8b}{3}$

[color=beige]. . . . . . . . . . [/color]$x \:=\:17\,-\,2b \,+\, \frac{1\,-\,2b}{3}$

$\text{Since }x\text{ is an integer, }1\,-\,2b\text{ be a multiple of 3.}$

$\text{The first time this happens is: }b = 2.
\;\;\;x \:=\:17\,-\,2(2)\,+_\,\frac{1\,-\,2(2)}{3} \;\;\;\Rightarrow\;\;\;x \,=\,12$

$\text{The next time this happens is: }b = 5.
\;\;\;x \:=\:17\,-\,2(5) \,+\,\frac{1\,-\,2(5)}{3} \;\;\;\Rightarrow\;\;\; \boxed{x \,=\,4}$

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