SOLUTIONS
you have to do is to place another I at the beginning and another 7
at the end—a considerable saving of labour ! These two, and the
example shown by the wizard, are the only two-figure multipliers,
but the number to be multiplied may always be increased. Thus, if
you prefix to 41096 the number 41095890, repeated any number of
times, the result may always be multiplied by 83 in the wizard's
peculiar manner.
83.—
The Ribbon Problem.
The solution is as follows : Place this rather lengthy number
on the ribbon, 02127659574468085106382978723404255319K
4393617. It may be multiplied by any number up to 46 inclu-
sive to give the same order of figures in the ring. The number
previously given can be multiplied by any number up to 16. I
made the limit 9 in order to put readers off the scent. The fact
is these two numbers are simply the recurring decimals that equal
one-seventeenth and one-forty-seventh respectively. Multiply the
one by seventeen and the other by forty-seven and you will get all
nines in each case.
84.—
The Japanese Ladies and the Carpet.
If the squares had not to be all the
same size, the carpet could be cut in four
pieces in any one of the three manners
shown. In each case the two pieces
marked A will fit together and form one
of the three squares, the other two
squares being entire. But in order to
have the squares exactly equal in size,
we shall require six pieces, as shown in
the larger diagram. No. 1 is a complete
square ; pieces 4 and 5 will form a
second square ; and pieces 2, 3, and 6 will form the third—all
of exactly the same size.
179
N
2
I £ .
U
ST
2
1 i
5
4
6^
3
