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Player A dealt some cards to player B and himself from a full pack of playing cards and laid the rest aside. A then said to B "If you give me a certain number of your cards, I will have four times as many cards as you will have. If I give you the same number of cards, I will have thrice as many cards as you will have." Of the given choices, which could represent the number of cards with player A?

(a) 9

(b) 31

(c) 12

(d) 35

My work :

$52 = y + (52 -y)$ -----initially y cards were dealt in total.

$y = c1 + (y-c1)$ -----cards with players A and B respectively

Condition 1 : $c1 + a = 4 ( y - c1 - a )$

Condition 2 : $c1 - b = 3 ( y - c1 + b)$

Eliminating y, we get c1 = 15a + 16b

The possible combinations for a and b are $(1,1), (1,2), (2,1)$.

So for the given choices, it should be the option b but I am not convinced. Is the approach correct?

piepi
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  • 52 can't be involved in the work, since the number of cards in the full deck is irrelevant. – Gerry Myerson Sep 23 '16 at 07:30
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    Actually if we denote $A$ as the number of cards of player $A$ and $B$ respectively the conditions say there is a number $n$, s.t. $A+n=4\cdot (B-n)$ and $A-n=3 \cdot (B+n)$. I am not quite sure what you mean with your $a$ and $b$ and you have quite a confusing notation about what denotes what. – ctst Sep 23 '16 at 07:37
  • y is the total number of cards currently in rotation. We give player A, c1 cards thus the player B gets y-c1 cards. a is the number of cards exchanged in first transaction, b in the second transaction. – piepi Sep 23 '16 at 07:42
  • I don't get you. I have put the first step just as a thought process, is that wrong? @GerryMyerson – piepi Sep 23 '16 at 07:44
  • @GerryMyerson No, the number of cards in the full deck is not irrelevant if we want to have a unique solutions (but you are right, if we can only pick our solution out of the 4 given numbers). You just need it to check that $A+B\leq 52$ holds (and then the solution is unique). – ctst Sep 23 '16 at 07:44
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    @piepi You have in your question: "If I give you the same number of cards, ..." hence you have to put $a=b$. Also you should really clean up your notations and first consider the important parts (e.g. how many cards each person gets and not how many cards are dealt, which is just the sum). This will help you in the future to focus on the important information. Despite this your approach is correct. – ctst Sep 23 '16 at 07:48
  • @ctst Ah shoot! Thanks. consider the important parts, how would I know what is important in a question without looking at all the details first, even the redundant ones? – piepi Sep 23 '16 at 07:52
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    @piepi By sticking first to the parts that are asked and given and then numbers you most of the time during your calculations: The numbers of cards dealt to $A$ is asked. The number of cards given between players, the number of cards dealt to $B$. You actually only need the sum of the dealt cards somewhere in the end (only if you don't have to choose your answer). Also that is what is called cleaning/improving/polishing your proof after you finished it (that is really important if you want anyone to ever look at your stuff!) :-) – ctst Sep 23 '16 at 08:11

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