As we were told that the man “succeeded” in carrying put his plan, we must try to find some loophole in the conditions.  He was to “enter every town once and only once,” and we find no prohibition against his entering once the town A after leaving it, especially as he has never left it since he was born, and would thus be “entering” it for the first time in his life.  But he must return at once from the first town he visits, and then he will have only 22 towns to visit, and as 22 is an even number, there is no reason why he should not end on the white square Z. A possible route for him is indicated by the dotted line from A to Z. This route is repeated by the dark lines in Fig. 1, and the reader will now have no difficulty in applying; it to the original map.  We have thus proved that the puzzle can only be solved by a return to A immediately after leaving it.

251.—­WATER, GAS, AND ELECTRICITY.

[Illustration]

According to the conditions, in the strict sense in which one at first understands them, there is no possible solution to this puzzle.  In such a dilemma one always has to look for some verbal quibble or trick.  If the owner of house A will allow the water company to run their pipe for house C through his property (and we are not bound to assume that he would object), then the difficulty is got over, as shown in our illustration.  It will be seen that the dotted line from W to C passes through house A, but no pipe ever crosses another pipe.

252.—­A PUZZLE FOR MOTORISTS.

[Illustration]

The routes taken by the eight drivers are shown in the illustration, where the dotted line roads are omitted to make the paths clearer to the eye.

253.—­A BANK HOLIDAY PUZZLE.

The simplest way is to write in the number of routes to all the towns in this manner.  Put a 1 on all the towns in the top row and in the first column.  Then the number of routes to any town will be the sum of the routes to the town immediately above and to the town immediately to the left.  Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc., in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with the other rows.  It will then be seen that the only town to which there are exactly 1,365 different routes is the twelfth town in the fifth row—­the one immediately over the letter E. This town was therefore the cyclist’s destination.

The general formula for the number of routes from one corner to the corner diagonally opposite on any such rectangular reticulated arrangement, under the conditions as to direction, is (m+n)!/m!n!, where m is the number of towns on one side, less one, and n the number on the other side, less one.  Our solution involves the case where there are 12 towns by 5.  Therefore m = 11 and n = 4.  Then the formula gives us the answer 1,365 as above.

254.—­ THE MOTOR-CAR TOUR.