[B] Mr. Oscar Hoppe, of New York, informs me that, after reading my statement in the Introduction, he was led to investigate the case of _n_ = 19, and after long and tedious work he succeeded in proving the number to be a prime. He submitted his proof to the London Mathematical Society, and a specially appointed committee of that body accepted the proof as final and conclusive. He refers me to the _Proceedings_ of the Society for 14th February 1918. As for the even values of _n_, the following curious series of factors will doubtless interest the reader. The numbers in brackets are primes. _n_ = 2 = (11) _n_ = 6 = (11) × 111 × 91 _n_ = 10 = (11) × 11,111 × (9,091) _n_ = 14 = (11) × 1,111,111 × (909,091) _n_ = 18 = (11) × 111,111,111 × 90,909,091 Or we may put the factors this way:-- _n_ = 2 = (11) _n_ = 6 = 111 × 1,001 _n_ = 10 = 11,111 × 100,001 _n_ = 14 = 1,111,111 × 10,000,001 _n_ = 18 = 111,111,111 × 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) × (101) _n_ = 8 = (11) × (101) × 10,001 _n_ = 12 = (11) × (101) × 100,010,001 _n_ = 16 = (11) × (101) × 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.