Trevor from Riccarton High School and Louise from Bourne Grammar School made excellent attempts at solving this puzzle. They both devised good methods for representing the new arrangement of the cards after each turn.
Shaun from Nottingham High School went a step further and realised that it was not necessary to number the cards just indicate whether a given position could or could not hold the chosen card.
Here's how he did that :
The cards are laid out in three columns, five rows.
An O denotes a card
the person could have possibly chosen, and X one we are certain they have
Here's the first arrangement :
Once the person tells us which column the card is in the first time, this
column is put in the middle, so no matter what column the card is in, the
cards are laid out like so:
The next time the cards are laid out there are three possibilities.
Possibility 1 is that the card is in the left column, possibility 2 that the card is in the centre column, and possibility 3 is that the card is in the right hand column.
When the correct column is put in the centre again, the cards will be laid out in one of the following ways:
Notice that whatever column the correct card is in, it is always in the
third row down. When this correct column is now put in between the two
other columns, the correct card becomes the eighth in the pack. This is
how the trick works.
Shaun then went on to consider the problem more generally :
For just 3 cards : The column only has to be given once, and the correct card can be
found, since there is only one card in each column.
For 6 cards : the correct card can be worked out on the second dealing of the
cards. The first deal will put it either on the top row of the right hand
column, or the bottom row of the left hand column. When the correct column
is then given by the person, the card will be known. However, it will not
be possible, like in the 15 card version, to say that it is the nth card
in the pack, since it resides on either of the two rows, and will be
either the third or fourth. Further repetitions of rearranging the cards
do nothing; the correct card stays in the same position.
For 9 cards the trick works exactly as with 15 cards, except only
two rearrangements, instead of three, are needed. This is because, after
the first rearrangement, the three possible correct cards are all put on
the same (middle) row, and so after the person has given the column of
their card a second time, it will be put into the middle of the pack, and
be the (n + 1)/2 th card.
For 12 cards the case is very similar to that for 6 cards, except one more
rearrangement is needed. The correct card will become stuck in one of two
possible positions, either left hand column, third row, or right hand
column, second row, allowing it to be discovered.
For 21 cards, intuition suggested to me that the number of cards required must be
an odd number multiplied by an odd number. Quite amazingly, the trick
still works, in the same amount of moves.
For 27 cards I attempted drawing diagrams like the ones above.
It still worked.
Then, for 33, the same trick was attempted. It now requires more
than three rearrangements to work, and so we can say the highest number of
cards the exact trick works for is 27.
This can be explained by looking at the position of possible correct
cards after the second rearrangement. In order for the trick to be
completed in three rearrangements, the number of rows the possible cards
take up can be no more than three. This is because when the correct
column is chosen next time, and the now possible correct cards sent into
the middle of the pack, they will take up more than one row, and so two or
more possibly correct cards will occupy the same column, and it will not be
possible to know for certain what the correct card is from the person's
next column declaration.
Also, the number of rows the possibly correct cards take up after the
first rearrangement must be an odd number, and have equal numbers of rows
which contain only cards which cannot possibly be the correct card in them
both above and below them, hence why the trick does not work for numbers of
cards made up of an even times and odd number. This is because the number
of possibly correct cards will be a third of the total cards. Because
they are in the middle of the pack, they will take up the middle row, and
then work their way outwards to those immediately above and below it.
However, as I said and explained above, they cannot extend to more than
three rows high in order to complete the trick in three rearrangements.
Well done Shaun