**THE TIRING IRONS.— solution**

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:—

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 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 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.

click here to go to my blog.

See more interesting puzzles at http://puzzles.50webs.org