Page:Amusements in mathematics.djvu/260

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 thefollowing 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:—

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 onethird of $(2^{n+1}-1)$.

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

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

It will be convenient to tabulate a few cases.

No. of Rings.	Total Positions	Positions used.	Positions. 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 off 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 risers 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 book-worm, in boring from the first to the last page