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Algebra

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27 Sep 2008 07:27:37 IST

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A stack of 2000 cards is labeled with the integers from 1 to 2000, with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card in the stack is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. This process -- placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack -- is repeated until all cards are on the table. It is found that, reading left to right, the labels on the cards are now in ascending order: 1,2,3,...,1999,2000. In the original stack of cards, how many cards were above the card labeled 1999?

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Brad

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Joined: 9 Sep 2007

Posts: 647

27 Sep 2008 07:36:28 IST

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Answer is 927 ......(I knew the solution before hand)

I don't think it would be fair to post it .

anchit saini

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Joined: 1 Feb 2008

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27 Sep 2008 09:08:58 IST

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I got it this way ->

Just going about it in the backward way solves it ,ie -

After all the cards are on the table , then ->

1 step -> We pick up a card from the table and place it on the top of the pile of cards

2 step -> Now , put the card from the bottom on the top

We find that 1999 comes on top in a stack of ->

3 , 6 ,12 , 24 ,..........., 1536

So the cards left are 2000 - 1536 = 464 on the table

Now , again continuing the procedure we place the card from the table on the top and then the bottom card on the top , until the last card is left which is placed on the top (In this step the bottom card is not place on the top , cos we need to have the original pile)

Hence , total cards above 1999 in original pile are 2*463 + 1 = 927

Rahul Duggal

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Joined: 10 May 2008

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27 Sep 2008 11:29:28 IST

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here is my solution

let 1^st card in the deck be denoted as card 1 irrespective of the integer possesed by it and so on

so beforehand we know integer on the 1^st card will be the predecessor of the number on the 3^rd card , no. on the 3^rd card will be the predecessor of the no. on the 5^th card and so on

so after the first 2000 operations on the initial deck of cards, we will obtain another deck of 1000 cards.

so after 1000 operations on the second deck of cards we obtain another deck containing 500 cards

again after 500 operations on the third deck of cards, we obtain another deck of card containing 250 cards

again after 250 operations on the deck of 250 cards we obtain another deck of 125 cards

now as the number of cards is odd, we have to decide that after 125 operations on the particular deck, how many cards will the resulting deck contain

as we are keeping the first card of every deck on the table, so we come to know that we will be keeping the 125^th on the table but we still dont know exactly how many cards will we be keeping on the table.

so by the general term of AP, we get 125=1+(n-1)2 so 125 will be the 63^rd card put on the table from this particular deck.

so as we have placed 63 cards from a deck of 125 cards, we obtain a deck of 62 cards.

now again after 62 operations on the deck of 62 cards, we obtain a deck of 31 cards.

as we proceeded in the deck of 125 cards, similarly we proceed in the deck of 31 cards.

thus after 31 operations on the deck of 31 cards, we obtain a deck containing 16 cards

again after 16 operations on the deck of 16 cards, we get a deck of 8 cards.

again after 8 operations on the deck of 8 cards, we get a deck of 4 cards.

again after 4 operations on the deck of 4 cards, we get a deck of 2 cards

again after 2 operations on the deck of 2 cards, we get a deck of 1 card

now after 1 operation on the last card, we have placed 2000 cards on the table.

as the cards placed on the table are in ascending order, we know that card with the integer 1999 on it will be the penultimate card. so we get to know that the particular card was the 1^st card of the deck containing 2 cards.

now proceeding backwards we get the first card of the deck of 2 cards is the 2^ndcard of the deck of 4 cards

now the 2^nd card of the deck of 4 cards is 4^th card of the deck of 8 cards

now 4^th card of the deck of 8 cards is the 8^thcard of the deck of 16 cards.

now as we move from the deck of 16 cards to the deck of 31 cards, we have to note that in the deck of 31 cards we kept the first card in the bottom of the deck and not on the table so the 1^st card of the deck of 16 cards is the 1^st card of the deck of 31 cards and 2^nd card of the deck of 16 is the 3^rd card of the deck of 31 so in general we can say n^th card of the deck of 16 cards is the (2n-1)^th card of the deck of 31, n>1

now the 8^th card of the deck of 16 cards is the 15^thcard of the deck of 31 cards.

similar as above the 16^thcard of the deck of 31 cards is the 29^thcard of the deck of 62 cards.

proceeding upwards we are putting the first card of the deck on the table so the 29^th card of the deck of 62 cards is the 58^th card of the deck of 125 cards.

now the 58^thcard of the deck of 125 cards is the 116^th card is the deck of 250 cards.

now the 116^thcard of the deck of 250 cards is 232^nd card of the deck of 500 cards.

now the 232^nd card of the deck of 500 cards is the 464^th card of the deck of 1000 cards.

finally the 464^th card of the deck of 1000 cards is the 928^th card of the deck of 2000 cards.

thus we can say there are 927 cards above the card with the number 1999 in the initial deck of 2000 cards

Rajat Sen

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Joined: 7 Aug 2007

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27 Sep 2008 20:09:17 IST

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Re:An interesting pne... everyone's invited

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