Take a whole pack, consisting of 52 cards, and desire some person in
company to draw out any card, at pleasure, without showing it. Having
assigned to the different cards their usual value, according to their
spots, call the knave 11, the queen 12, and the king 13. Then add the
spots of the first card to those of the second; the last sum to the
third; and so on, always rejecting 13, and keeping the remainder to add
to the following card. It may be readily seen that it is needless to
reckon the kings, which are counted 13. If any spots remain at the last
card, you must abstract them from 13, and the remainder will indicate
the card that has been drawn: if 12 remains, it has been an ace; but if
nothing remains, it has been a king.
_Demonstration._—Since a complete pack contains 13 cards of each suit,
the values of which are 1, 2, 3, &c., as far as 13, the sum of all the
spots of each of the different suits will be 7 times 13 (91), which is a
multiple of 13; consequently the quadruple is also a multiple of 13: if
we add the spots of all the cards, always rejecting 13, the remainder at
last must be 0. Hence it is evident, that if a card, the spots of which
are less than 13, be drawn, the difference between its spots and 13 will
be what is wanting to complete the number. If, at the end, then, instead
of attaining to 13, we attain only to 10, for example, it is plain, that
the card wanting is a 3; and if we attain exactly to 13, the card
missing must be equivalent to 13; that is, it must be a king.