# Amusements in Mathematics

SOLUTIONS 1-100
1.—A POST-OFFICE PERPLEXITY.—solution
The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8 twopence-
halfpenny stamps, which delivery exactly fulfils the conditions and represents a cost of
five shillings.
2.—YOUTHFUL PRECOCITY.—solution
The price of the banana must have been one penny farthing. Thus, 960 bananas would
cost £5, and 480 sixpences would buy 2,304 bananas.
3.—AT A CATTLE MARKET.—solution
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must
have taken 21. There were thus 39 animals altogether.
4.—THE BEANFEAST PUZZLE.—solution
The cobblers spent 35s., the tailors spent also 35s., the hatters spent 42s., and the glovers
spent 21s. Thus, they spent altogether £6,13s., while it will be found that the five cobblers
spent as much as four tailors, twelve tailors as much as nine hatters, and six hatters as
much as eight glovers.
5.—A QUEER COINCIDENCE.—solution
Puzzles of this class are generally solved in the old books by the tedious process of
"working backwards." But a simple general solution is as follows: If there are n players,
the amount held by every player at the end will be m(2n), the last winner must have held
m(n+1) at the start, the next m(2n+1), the next m(4n+1), the next m(8n+1), and so on to
the first player, who must have held m(2n-1n+1).
Thus, in this case, n = 7, and the amount held by every player at the end was 27 farthings.
Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with
113, B with 225, and A with 449 farthings.