# A Level Maths Problem

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I have a question and i can see how to solve it, but i can't see how to do it algebraically. If anyone can help or point me in the right direction i would be grateful:

I have a pack of playing cards, numbered 1-52. They are in order with 1 on the top and 52 on the bottom. I go through the pack discarding every other card, so I end up with a new pile which has 1 on top, then 3, 5, 7.... up to 51. I do the same with the new pile, and so on, getting even smaller piles.

NOTE: If the last card in the pile is one i keep, I will discard the top card of the subsequent pile.

What is the last card i'm left with?

What if i do it again, but discard the 1, keep the 2m etc...?

What if i start with 100 cards?

I have a pack of playing cards, numbered 1-52. They are in order with 1 on the top and 52 on the bottom. I go through the pack discarding every other card, so I end up with a new pile which has 1 on top, then 3, 5, 7.... up to 51. I do the same with the new pile, and so on, getting even smaller piles.

NOTE: If the last card in the pile is one i keep, I will discard the top card of the subsequent pile.

What is the last card i'm left with?

What if i do it again, but discard the 1, keep the 2m etc...?

What if i start with 100 cards?

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#2

(Original post by

I have a question and i can see how to solve it, but i can't see how to do it algebraically. If anyone can help or point me in the right direction i would be grateful:

I have a pack of playing cards, numbered 1-52. They are in order with 1 on the top and 52 on the bottom. I go through the pack discarding every other card, so I end up with a new pile which has 1 on top, then 3, 5, 7.... up to 51. I do the same with the new pile, and so on, getting even smaller piles.

NOTE: If the last card in the pile is one i keep, I will discard the top card of the subsequent pile.

What is the last card i'm left with?

What if i do it again, but discard the 1, keep the 2m etc...?

What if i start with 100 cards?

**prajeetboi**)I have a question and i can see how to solve it, but i can't see how to do it algebraically. If anyone can help or point me in the right direction i would be grateful:

I have a pack of playing cards, numbered 1-52. They are in order with 1 on the top and 52 on the bottom. I go through the pack discarding every other card, so I end up with a new pile which has 1 on top, then 3, 5, 7.... up to 51. I do the same with the new pile, and so on, getting even smaller piles.

NOTE: If the last card in the pile is one i keep, I will discard the top card of the subsequent pile.

What is the last card i'm left with?

What if i do it again, but discard the 1, keep the 2m etc...?

What if i start with 100 cards?

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(Original post by

Can you post the original question and your thoughts about how you solved it?

**mqb2766**)Can you post the original question and your thoughts about how you solved it?

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#4

(Original post by

that is the original question, and i only solved the first question of the 3 but literally doing it lol

**prajeetboi**)that is the original question, and i only solved the first question of the 3 but literally doing it lol

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(Original post by

Ok, if you're solving the question, I won't spoil your fun.

**mqb2766**)Ok, if you're solving the question, I won't spoil your fun.

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#6

(Original post by

no i need help pls idk what to do

**prajeetboi**)no i need help pls idk what to do

If you're really stuck, describe which part.

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#7

(Original post by

Ok, if you're solving the question, I won't spoil your fun.

**mqb2766**)Ok, if you're solving the question, I won't spoil your fun.

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(Original post by

I think they meant they worked out the answer to the first part by literally doing it with cards.

**GabiAbi84**)I think they meant they worked out the answer to the first part by literally doing it with cards.

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(Original post by

Which part and what are your thoughts? If you don't know how to proceeed, maybe simplify the problem and work out what happens for a small number of cards.

If you're really stuck, describe which part.

**mqb2766**)Which part and what are your thoughts? If you don't know how to proceeed, maybe simplify the problem and work out what happens for a small number of cards.

If you're really stuck, describe which part.

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#10

(Original post by

i want to do the whole thing properly pls help

**prajeetboi**)i want to do the whole thing properly pls help

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(Original post by

The forum rules are to help, so pls upload what you've done / and describe what you're confused about

**mqb2766**)The forum rules are to help, so pls upload what you've done / and describe what you're confused about

I have only got as far as doing the first question and i did it by writing the numbers out and eliminating them, until i was left with 41.

I want to do all 3 of the questions algebraically rather than like this so please help me out thanks

mqb2766

Last edited by prajeetboi; 3 weeks ago

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#12

(Original post by

ok, i cant really take a pic so i'll just tell you.

I have only got as far as doing the first question and i did it by writing the numbers out and eliminating them, until i was left with 41.

I want to do all 3 of the questions algebraically rather than like this so please help me out thanks

mqb2766

**prajeetboi**)ok, i cant really take a pic so i'll just tell you.

I have only got as far as doing the first question and i did it by writing the numbers out and eliminating them, until i was left with 41.

I want to do all 3 of the questions algebraically rather than like this so please help me out thanks

mqb2766

Let us firstly consider what happens if we start with an even/odd number of cards, and if we begin by discarding the first/second card.

Let be the number of cards we begin with.

__Case 1: n is even, and we begin by discarding the second card__

In this scenario, we will discard every even card, therefore the final card will be discarded as well. This hence does not imply that on the next sweep we discard the first card.

So out of the pile numbered {1,2,3,4,5,...,n} we end up with a pile numbered {1,3,5,...,n-1}.

We end up with cards of the form for , which represents our new card order.

If n/2 is even, then we do the same thing and end up with {1,5,9,...,n-2}. Note that is odd here, so we need to ensure that for . Hence, the surviving numbers are of the form .

Note that we can deduce if is divisible by , then the resulting surviving numbers are consecutive sweeps are of the form for .

__Case 2: n is odd, and we begin by discarding the second card__

In this scenario, we discard every second card but keep the final card. This hence means that on the following sweep we begin by discarding the first card.

Out of the cards numbered {1,2,3,4,5,...,n-1,n} we obtain the list {1,3,5,7,...,n}

We end up with number of cards, and they are all of the form for .

__Case 3: n is even, and we begin by discarding the first card__

In this case, we discard the first, third, etc... card and keep the final card. This hence means that on the following sweep we begin by discarding the first card.

Out of the cards numbered {1,2,3,4,5,...,n-1,n} we obtain {2,4,6,8,...,n-4,n-2,n}.

The numbers we keep are of the form where .

If this number of cards is even again, we go again and obtain {4,8,...,n-4,n}

and the numbers are of the form for .

We can generalise and see that if is divisible by , then after consecutive sweeps the surviving numbers are of the form for .

__Case 4: n is odd, and we begin by discarding the first card__

Here it is the same game as in Case 3 except the discard the final card.

From {1,2,3,4,5,6,...,n} we are left with {2,4,6,...,n-1}.

The leftover numbers are of the form for .

So, to get the answer for 52 cards, we deal with this systematically.

52 is div by 4 therefore after two consecutive sweeps we obtain numbers of the form

for .

We have 13 cards remaining, so we activate Case 2. The surviving numbers are of the form

and then we are left with 7 cards, and we need to begin by picking the first card [Case 4]. The surviving numbers are

for .

So we have 3 cards. We go back to Case 2. The surviving numbers are of the form

and we have 2 cards remaining. Case 3 yields the surviving card as

Thus, the surviving card is .

Anyway, I got to run now, so hopefully you can use this to check the 100 case.

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#13

(Original post by

Deducing the algebraic approach in this problem is like opening can of worms.

**RDKGames**)Deducing the algebraic approach in this problem is like opening can of worms.

This isn't my own work, but unless I'm stealing an approach to the wrong problem, then then there's actually a fairly straightforward way of solving this.

First, solve the case where the number of cards is a power of 2.

Next, reduce the case where the number of cards is NOT a power of 2 to the previous case.

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