s
will doubtless interest the reader. The numbers in brackets are primes.

_n_ = 2 = (11)

_n_ = 6 = (11) x 111 x 91

_n_ = 10 = (11) x 11,111 x (9,091)

_n_ = 14 = (11) x 1,111,111 x (909,091)

_n_ = 18 = (11) x 111,111,111 x 90,909,091

Or we may put the factors this way:--

_n_ = 2 = (11)

_n_ = 6 = 111 x 1,001

_n_ = 10 = 11,111 x 100,001

_n_ = 14 = 1,111,111 x 10,000,001

_n_ = 18 = 111,111,111 x 1,000,000,001

In the above two tables _n_ is of the form 4_m_ + 2. When _n_ is of the
form 4_m_ the factors may be written down as follows:--

_n_= 4 = (11) x (101)

_n_ = 8 = (11) x (101) x 10,001

_n_ = 12 = (11) x (101) x 100,010,001

_n_ = 16 = (11) x (101) x 1,000,100,010,001.

When _n_ = 2, we have the prime number 11; when _n_ = 3, the factors are
3 . 37; when _n_ = 6, they are 11 . 3 . 37 . 7. 13; when _n_ = 9, they
are 3^2 . 37 . 333,667. Therefore we know that factors of _n_ = 18 are
11. 3^2 . 37 . 7 . 13 . 333,667, while the remaining factor is composite
and can be split into 19 . 52579. This will show how the working may be
simplified when _n_ is not prime.

48.--_The Riddle of the Frogs' Ring._

The fewest possible moves in which this puzzle can be solved are 118. I
will give the complete solution. The black figures on white discs move
in the directions of the hands of a clock, and the white figures on black
discs the other way. The following are the numbers in the order in which
they move. Whether you have to make a simple move or a leaping move will
be clear from the position, as you never can have an alternative. The
moves enclosed in brackets are to be played five times over: 6, 7, 8, 6,
5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6,
5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2,
12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We
thus have made 118 moves within the conditions, the black frogs have
changed places with the white ones, and 1 and 12 are side by side in the
positions stipulated.

The general solution in the case of this puzzle is 3_n_^{2} + 2_n_ - 2
moves, where the number of frogs of each colour is _n_. The law governing
the sequence of moves is easily discovered by an examination of the
simpler cases, where _n_ = 2, 3, and 4.

If, instead of 11 and 12 changing places, the 6 and 7 must interchange,
the expression is _n_^{2} + 4_n_ + 2 moves. If we give _n_ the value 6,
as in the exam