The performer lays upon the table ten cards, side by side, face
downwards. Anyone is then at liberty (the performer meanwhile
retiring from the room) to shift any number of the cards (from one
to nine inclusive) from the right hand end of the row to the left,
but retaining the order of the cards so shifted. The performer, on
his return, makes a little speech: “Ladies and gentlemen, you have
shifted a certain number of these cards. Now, I don’t intend to ask
you a single question. By a simple mental calculation I can ascertain
the number you have moved, and by my clairvoyant faculty, though the
cards are face downwards, I shall pick out one corresponding with
that number. Let me see” (pretends to calculate, and presently turns
up a card representing “five”). “You shifted five cards and I have
turned up a five, the exact number.”
The cards moved are not replaced, but the performer again retires,
and a second person is invited to move a few more from right to
left. Again the performer on his return takes up the correct card
indicating the number shifted. The trick, unlike most others, may be
repeated without fear of detection.
[Illustration]
The principle is arithmetical. To begin with, the cards are arranged,
unknown to the spectators, in the following order:
Ten, nine, eight, seven, six, five, four, three, two, one.
Such being the case, it will be found that, however many are shifted
from right to left, the _first_ card of the new row will indicate
their number. Thus, suppose _three_ are shifted. The new order of the
cards will then be:
_Three_, two, one, ten, nine, eight, seven, six, five, four.
So far, the trick is easy enough, but the method of its continuance
is a trifle more complicated. To tell the position of the indicating
card after the second removal, the performer privately adds the
number of that last turned up (in this case _three_) to its place in
the row--_one_. That gives us _four_, the card to be turned up
after the next shift will be the fourth. Thus, suppose six cards are now
shifted, their new order will be:
Nine, eight, seven, _six_, five, four, three, two, one, ten.
Had five cards only been shifted, the _five_ would have been
fourth in the row, and so on.
The performer now adds _six_, the number of the card, to its place
in the row, _four_: the total, _ten_, gives him the position of
the indicator for the next attempt. Thus, suppose four cards are next
shifted, the new order will be:
Three, two, one, ten, nine, eight, seven, six, five, _four_.
The next calculation, 4 and 10, gives us a total 14. The ten is, in
this case, cancelled, and the fourteen regarded as _four_, which
will be found to be the correct indicator for the next shifting.
It looks more mystifying if the performer be blindfolded, for he can
tell the position of the cards with his fingers. Keeping his hand
on the card, he asks, “Will you please tell me how many cards were
shifted?” As soon as the answer is given, he exhibits the card, and
can continue the trick as long as he pleases.
5. Find 16 numbers in arithmetical progression (common difference 2)
whose sum shall be equal to 7552, and arrange them in 4 columns, 4
numbers in each column--or, in other words, arrange in a square of 16
numbers that when added vertically, horizontally, or diagonally, the
sum of each 4 numbers will amount to 1888.