408 Answers4 D., 2 S., ace H., ace c., 7 S., 5 D., 9 H., 2 H., Jack S., 6 D., Queen c., 6 c.,
10 H., 3 S., 3 H., 7 D., 4 c., 2 c., 8 S., Jack H., 4 H., 8 D., Jack c., 4 S., Queen
S., King C., 9 D., 5 H., 10 c., Queen H., 10 D., 9 S., 6 H., 5 S.
[All such arrangements for spelling cards can be solved quickly by starting
with the last card to be spelled, then performing all the required operations in
reverse order, finishing with the full pack, or packet, of cards.-M. G.]
- CARD SHUFFLING
To shuffle fourteen cards in the manner described, so that the cards shall
return to their original order, requires fourteen shuffles, though with sixteen
cards we require only 5. We cannot go into the law of the thing, but the reader
will find it an interesting investigation.
[For the mathematical theory of this shuffle, see W. W. Rouse Ball,
Mathematical Recreations and Essays, revised eleventh edition, edited by
H. S. M. Coxeter (Macmillan, 1960), pp. 310-311. The shuffle is sometimes
called "Monge's shuffle" after Gaspard Monge, a famous eighteenth century
French mathematician who was the first to investigate it.-M. G.]
- A CHAIN PUZZLE
To open a link and join it again will cost 3¢. By opening one link at the end
of each of the thirteen pieces the cost will be 39¢, so it would be cheaper than
that to buy a new chain. If there happened to be a piece of twelve links, all
these twelve coUld be opened to join the remaining twelve pieces at a cost also
of 36¢. If there had happened to be two pieces together, containing eleven
links, all these could be opened to join the remaining eleven pieces at a cost
of 33¢.
The best that can be done is to open three pieces containing together ten
links to join the remaining ten pieces at a cost of 30¢. This is possible if we
break up the piece of four links and two pieces of three links. Thus, if we in-
clude the piece of three links that was shown in the middle row as one of the
three link pieces, we shall get altogether five large links and five small ones.
If we had been able to find four pieces containing together nine links
we should save another 3¢, but this is not possible, nor can we find five pieces
containing together eight links, and so on, therefore the correct answer is as
stated, 30¢.