Part 110

413.--A CHESSBOARD FALLACY.

The explanation of this little fallacy is as follows. The error lies in assuming that the little triangular piece, marked C, is exactly the same height as one of the little squares of the board. As a matter of fact, its height (if we make the sixty-four squares each a square inch) will be 1+1/7 in. Consequently the rectangle is really 9+1/7 in. by 7 in., so that the area is sixty-four square inches in either case. Now, although the pieces do fit together exactly to form the perfect rectangle, yet the directions of the horizontal lines in the pieces will not coincide. The new diagram above will make everything quite clear to the reader.

414.--WHO WAS FIRST?

Biggs, who saw the smoke, would be first; Carpenter, who saw the bullet strike the water, would be second; and Anderson, who heard the report, would be last of all.

415.--A WONDERFUL VILLAGE.

When the sun is in the horizon of any place (whether in Japan or elsewhere), he is the length of half the earth's diameter more distant from that place than in his meridian at noon. As the earth's semi-diameter is nearly 4,000 miles, the sun must be considerably more than 3,000 miles nearer at noon than at his rising, there being no valley even the hundredth part of 1,000 miles deep.

416.--A CALENDAR PUZZLE.

The first day of a century can never fall on a Sunday; nor on a Wednesday or a Friday.

417.--THE TIRING-IRONS.

I will give my complete working of the solution, so that readers may see how easy it is when you know how to proceed. And first of all, as there is an even number of rings, I will say that they may all be taken off in one-third of (2^(n + 1) - 2) moves; and since n in our case is 14, all the rings may be taken off in 10,922 moves. Then I say 10,922 - 9,999 = 923, and proceed to find the position when only 923 out of the 10,922 moves remain to be made. Here is the curious method of doing this. It is based on the binary scale method used by Monsieur L. Gros, for an account of which see W.W. Rouse Ball's Mathematical Recreations.

Divide 923 by 2, and we get 461 and the remainder 1; divide 461 by 2, and we get 230 and the remainder 1; divide 230 by 2, and we get 115 and the remainder nought. Keep on dividing by 2 in this way as long as possible, and all the remainders will be found to be 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, the last remainder being to the left and the first remainder to the right. As there are fourteen rings and only ten figures, we place the difference, in the form of four noughts, in brackets to the left, and bracket all those figures that repeat a figure on their left. Then we get the following arrangement: (0 0 0 0) 1 (1 1) 0 (0) 1 (1) 0 1 (1). This is the correct answer to the puzzle, for if we now place rings below the line to represent the figures in brackets and rings on the line for the other figures, we get the solution in the required form, as below:--

O O O OO ------------------------- OOOO OO O O O

This is the exact position of the rings after the 9,999th move has been made, and the reader will find that the method shown will solve any similar question, no matter how many rings are on the tiring-irons. But in working the inverse process, where you are required to ascertain the number of moves necessary in order to reach a given position of the rings, the rule will require a little modification, because it does not necessarily follow that the position is one that is actually reached in course of taking off all the rings on the irons, as the reader will presently see. I will here state that where the total number of rings is odd the number of moves required to take them all off is one-third of (2^(n + 1) - 1).

With n rings (where n is odd) there are 2^n positions counting all on and all off. In (1/3)(2^(n + 1) + 2) positions they are all removed. The number of positions not used is (1/3)(2^n - 2).

With n rings (where n is even) there are 2^n positions counting all on and all off. In (2^(n + 1) + 1) positions they are all removed. The number of positions not used is here (1/3)(2^n - 1).

It will be convenient to tabulate a few cases.

+--------+------------+-----------+-----------+ | No. of | Total | Positions | Positions | | Rings. | Positions. | used. | not used. | +--------+------------+-----------+-----------+ | 1 | 2 | 2 | 0 | | 3 | 8 | 6 | 2 | | 5 | 32 | 22 | 10 | | 7 | 128 | 86 | 42 | | 9 | 512 | 342 | 170 | | | | | | | 2 | 4 | 3 | 1 | | 4 | 16 | 11 | 5 | | 6 | 64 | 43 | 21 | | 8 | 256 | 171 | 85 | | 10 | 1024 | 683 | 341 | +--------+------------+-----------+-----------+

Note first that the number of positions used is one more than the number of moves required to take all the rings off, because we are including "all on" which is a position but not a move. Then note that the number of positions not used is the same as the number of moves used to take off a set that has one ring fewer. For example, it takes 85 moves to remove 7 rings, and the 42 positions not used are exactly the number of moves required to take off a set of 6 rings. The fact is that if there are 7 rings and you take off the first 6, and then wish to remove the 7th ring, there is no course open to you but to reverse all those 42 moves that never ought to have been made. In other words, you must replace all the 7 rings on the loop and start afresh! You ought first to have taken off 5 rings, to do which you should have taken off 3 rings, and previously to that 1 ring. To take off 6 you first remove 2 and then 4 rings.

418.--SUCH A GETTING UPSTAIRS.

Number the treads in regular order upwards, 1 to 8. Then proceed as follows: 1 (step back to floor), 1, 2, 3 (2), 3, 4, 5 (4), 5, 6, 7 (6), 7, 8, landing (8), landing. The steps in brackets are taken in a backward direction. It will thus be seen that by returning to the floor after the first step, and then always going three steps forward for one step backward, we perform the required feat in nineteen steps.

419.--THE FIVE PENNIES.

First lay three of the pennies in the way shown in Fig. 1. Now hold the remaining two pennies in the position shown in Fig. 2, so that they touch one another at the top, and at the base are in contact with the three horizontally placed coins. Then the five pennies will be equidistant, for every penny will touch every other penny.

420.--THE INDUSTRIOUS BOOKWORM.

The hasty reader will assume that the bookworm, in boring from the first to the last page of a book in three volumes, standing in their proper order on the shelves, has to go through all three volumes and four covers. This, in our case, would mean a distance of 91/2 in., which is a long way from the correct answer. You will find, on examining any three consecutive volumes on your shelves, that the first page of Vol. I. and the last page of Vol. III. are actually the pages that are nearest to Vol. II., so that the worm would only have to penetrate four covers (together, 1/2 in.) and the leaves in the second volume (3 in.), or a distance of 31/2 inches, in order to tunnel from the first page to the last.

421.--A CHAIN PUZZLE.

To open and rejoin a link costs threepence. Therefore to join the nine pieces into an endless chain would cost 2s. 3d., whereas a new chain would cost 2s. 2d. But if we break up the piece of eight links, these eight will join together the remaining eight pieces at a cost of 2s. But there is a subtle way of even improving on this. Break up the two pieces containing three and four links respectively, and these seven will join together the remaining seven pieces at a cost of only 1s. 9d.

422.--THE SABBATH PUZZLE.

The way the author of the old poser proposed to solve the difficulty was as follows: From the Jew's abode let the Christian and the Turk set out on a tour round the globe, the Christian going due east and the Turk due west. Readers of Edgar Allan Poe's story, Three Sundays in a Week, or of Jules Verne's Round the World in Eighty Days, will know that such a proceeding will result in the Christian's gaining a day and in the Turk's losing a day, so that when they meet again at the house of the Jew their reckoning will agree with his, and all three may keep their Sabbath on the same day. The correctness of this answer, of course, depends on the popular notion as to the definition of a day--the average duration between successive sun-rises. It is an old quibble, and quite sound enough for puzzle purposes. Strictly speaking, the two travellers ought to change their reckonings on passing the 180th meridian; otherwise we have to admit that at the North or South Pole there would only be one Sabbath in seven years.