I first gave this problem for six persons on ten days, in the _Daily Mail_ for the 13th and 16th October 1905, and it has since been discussed in various periodicals by mathematicians. Of course, it is easily seen that the maximum number of sittings for _n_ persons is (_n_ - 1)(_n_ -2)/2 ways. The comparatively easy method for solving all cases where _n_ is a prime+1 was first discovered by Ernest Bergholt. I then pointed out the form and construction of a solution that I had obtained for 10 persons, from which E. D. Bewley found a general method for all even numbers. The odd numbers, however, are extremely difficult, and for a long time no progress could be made with their solution, the only numbers that could be worked being 7 (given above) and 5, 9, 17, and 33, these last four being all powers of 2+1. At last, however (though not without much difficulty), I discovered a subtle method for solving all cases, and have written out schedules for every number up to 25 inclusive. The case of 11 has been solved also by W. Nash. Perhaps the reader will like to try his hand at 13. He will find it an extraordinarily hard nut. The solutions for all cases up to 12 inclusive are given in _A. in M._, pp. 205, 206. 91.--_The Five Tea Tins._ There are twelve ways of arranging the boxes without considering the pictures. If the thirty pictures were all different the answer would be 93,312. But the necessary deductions for cases where changes of boxes may be made without affecting the order of pictures amount to 1,728, and the boxes may therefore be arranged, in accordance with the conditions, in 91,584 different ways. I will leave my readers to discover for themselves how the figures are to be arrived at. 92.--_The Four Porkers._ The number of ways in which the four pigs may be placed in the thirty-six sties in accordance with the conditions is seventeen, including the example that I gave, not counting the reversals and reflections of these arrangements as different. Jaenisch, in his _Analyse Mathématique au jeu des Échecs_ (1862), quotes the statement that there are just twenty-one solutions to the little problem on which this puzzle is based. As I had myself only recorded seventeen, I examined the matter again, and found that he was in error, and, doubtless, had mistaken reversals for different arrangements. Here are the seventeen answers. The figures indicate the rows, and their positions show the columns. Thus, 104603 means that we place a pig in the first row of the _first_ column, in no row of the _second_ column, in the fourth row of the _third_ column, in the sixth row of the _fourth_ column, in no row of the _fifth_ column, and in the third row of the _sixth_ column. The arrangement E is that which I gave in diagram form:-- A. 104603 B. 136002 C. 140502 D. 140520 E. 160025 F. 160304 G. 201405 H. 201605 I. 205104 J. 206104 K. 241005 L. 250014 M. 250630 N. 260015 O. 261005 P. 261040 Q. 306104