Counting Cards
Problem
A magician took a suit of thirteen cards and held them in his hand face down.
He took the top card off the pile and put it at the bottom , saying 'A' as he did it. He took the next card and said 'C' as he put it at the bottom. He took the next card and turned it over, saying 'E' as he did it - and the card was an ACE!
He carried on with the letters T-W-O and as he said the 'O' he turned over the card and it was a TWO!
He carried on with T-H-R-E-E, the FOUR and so on, and in each case as he said the last letter of the name he turned over the card and everyone was amazed that he had predicted what it would be.
How did this work? You might like to see our NRICH magician Charlie demonstrating this in the video below.
Try with a slightly simpler version: Starting with ten cards numbered $1$ to $10$, can you arrange them in such a way that - starting with the arranged pile face down - you can spell out each card and reveal it as you announce its last letter?
Can you explain a way of doing this systematically so that you can quickly arrange any number of cards to make the trick work?
And the really hard bit: What would happen if you counted the number of cards equal to the value of the next card (so, if the next card was due to be a six - you would put five cards on the bottom of the pack and reveal the sixth)?
Getting Started
You need to be systematic.
Try with a smaller number of cards first or using two sets of cards one to help you keep your "magic" cards in the correct positions.
Student Solutions
Maggie of St Anne's School sent the following partial solution, well done Maggie.
You have to arrange the cards in a certain order . For example,
the order for 1 2 3 4 5 would be 2 4 1 3 5
O...N...E
1...2...3
T...W...O
4...5...1
T...H...R...E...E
2...4...5...2...4
F...O...U...R
5...2...5...2...
F...I...V...E
5...5...5...5
So the cards would be arranged with
Ace in position 3
Two in position 1
Three in position 4
Four in position 2
Five in position 5
There were a number of other anonymous solutions and a solution from Andrei of School 205 Bucharest. Here is one method suggested by one person:
Let's say you have 10 cards - imagine 10 positions that represent the order of the cards in the pack
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Then you can place each card in turn in each position - using up the space so:
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
O
|
N
|
1
|
T
|
W
|
2
|
T
|
H
|
R
|
E
|
but you run out of room before you can spell out THREE so you go back to the beginning and 3 will go in position 1. The you start spelling FOUR but you have to jump over position 3 because it has a 1 in it and position 6 because it has a 2 in it, so 4 ends up in position 7:
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
3
|
F
|
1
|
O
|
U
|
2
|
4
|
So you end up with:
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
3
|
5
|
1
|
8
|
10
|
2
|
4
|
6
|
7
|
9
|
This method works for any number of cards and whether you use their names or their values.
Teachers' Resources
Counting Cards
How did this work?
Try with a slightly simpler version: Starting with ten cards numbered $1$ to $10$, can you arrange them in such a way that - starting with the arranged pile face down - you can spell out each card and reveal it as you announce its last letter?
Can you explain a way of doing this systematically so that you can quickly arrange any number of cards to make the trick work?
And the really hard bit: What would happen if you counted the number of cards equal to the value of the next card (so, if the next card was due to be a six - you would put five cards on the bottom of the pack and reveal the sixth)?
Why do this problem?
Possible approach
Key questions
Possible extension
Possible support
Handouts for teachers are available here (word document, pdf document), with the problem on one side and the notes on the other.