One Out One Under
Imagine a stack of numbered cards with one on top. Discard the top,
put the next card to the bottom and repeat continuously. Can you
predict the last card?
Problem
Image
Imagine a stack of cards, numbered one to seven in order, with one on the top.
Discard the top card (card 1) and put the next card (card 2) to the bottom of the stack.
Put out card 3 and then put card 4 under.
Continue putting one out and one under until you are holding only one card - it should be card 6.
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What happens if you start with 8 cards ?
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What happens if you start with 9 cards ?
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How much can you find out about this situation ?
Send us whatever you discover.
Continue exploring until you can predict the final card and explain why your method works.
Getting Started
For the first part, try starting with only four cards and compare the resulting final card with the final card when you start with eight cards.
Notice anything ?
What might be a good number of cards to try next ?
When you are ready to look for a general result, be systematic and list your results, for example : seven cards and the final card is 6, eight cards and the final card is . . . , nine cards . . . ., ten cards . . .
Remember to collect results in the other direction away from seven cards. Six cards and the final card was . . . etc.
If you've been careful you will see a clear pattern.
First describe the pattern. Then try to account for it.
Student Solutions
Thank you to Stacey from Wales High School, Dmitri from Cork in Ireland, and many others for ideas about what's going on here.
If you start with 8 cards you end up with the number 8, and if you start with 9 cards you end up with the number 2.
If you start with 2 or 4 or 8 or 16 the last card is 2, 4, 8 or 16 to match.
After any one of those, for example 9 after 8, the last card moves on by 2.
So 8 cards finishes with 8, 9 cards finishes with 2, 10 cards finishes with 4, 11 cards finishes with 6, and so on until 14 finishes with 12, 15 finishes with 14 and 16 (the next power of 2 after 8) finishes with 16.
Then it all happens again, in the same way : 17 cards finishes with 2, 18 cards with 4 and so on.
Here's why that happens
For example starting with : 1 2 3 4 5 6 7 8Every second card is kept and we get : 2 4 6 8
Half the cards have gone, the second card of each pair.
The same thing happens and we are left with: 4 8
Half the cards have been lost again, as before the second card in each pair has gone.
Finally 4 8 goes down to 8
It's always the second card of the pair that stays in.
So when the number of cards is a plain power of 2, like 8, only half the cards stay, then only half of those, and so on until it's just the end card, like 8, that remains.
Now for the other numbers :
Start at a plain power of two, like 8, and increase by 1, that's 9, and make the first move of 'one out and one under'.Now we have 8 cards again, we had 9 but one's gone out.
And for 8 we know what will happen, we'll be left with the last card in that order as the final card remaining.
So what's the order ?
We've done one out and one under so all positions will have moved by two cards.
So we don't finish on the 8, or the next card (1 - it's out), but the one after that, that's the 2.
Now think about 10 cards.
Make the first move : 'one out and one under'.
And then there are 9 cards and we know that that will finish with 2, but all the positions have moved by 2 places, so the final card is the 4.
The same thing happens again and again - when there's one more card at the start the last card moves 2 on from the previous result.
Nice reasoning !
Here is a table sent in to us by children at Weston Turville CE School:
Start Card | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
Card Left | 2 | 2 | 4 | 2 |
4
|
6 | 8 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
Start Card | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 |
Card Left | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32 |
In the table shown you can see that every time that you have a number of cards to start with that is a power of $2$ i.e. $\{2, 2^2, 2^3, 2^4, 2^5, 2^6\}$ then it results in itself.
Teachers' Resources
This is a three star problem so hopefully it will keep students thinking and testing their reasoning for some while.
Working systematically helps greatly. A table of results shows a very clear pattern, and from there the challenge is to find a way of accounting for that pattern.
This problem offers an excellent opportunity for reasoning from the n case to the n + 1 case.
Seeing the n case result embedded within the n + 1 starting line up.
You may be interested to know that this problem was taken from the book called Maths trails - Visualising