Part 30

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 4m + 2. When n is of the form 4m 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 3n^{2} + 2n - 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} + 4n + 2 moves. If we give n the value 6, as in the example of the Frogs' Ring, the number of moves would be 62.

For a general solution of the case where frogs of one colour reverse their order, leaving the blank space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" in A. in M., p. 193.

THE STRANGE ESCAPE OF THE KING'S JESTER

Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.

49.--The Mysterious Rope.

When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.

50.--The Underground Maze.

How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side--one piece like a U, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.

See the article, "Mazes, and How to Thread Them," in A. in M.

51.--The Secret Lock.

This puzzle entailed the finding of an English word of three letters, each letter being found on a different dial. Now, there is no English word composed of consonants alone, and the only vowel appearing anywhere on the dials is Y. No English word begins with Y and has the two other letters consonants, and all the words of three letters ending in Y (with two consonants) either begin with an S or have H, L, or R as their second letter. But these four consonants do not appear. Therefore Y must occur in the middle, and the only word that I can find is "PYX," and there can be little doubt that this was the word. At any rate, it solves our puzzle.

52.--Crossing the Moat.