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.