271.—FIFTEEN LETTER PUZZLE.
The following will be found to comply with the conditions of grouping:—
| ALE | MET | MOP | BLM |
| BAG | CAP | YOU | CLT |
| IRE | OIL | LUG | LNR |
| NAY | BIT | BUN | BPR |
| AIM | BEY | RUM | GMY |
| OAR | GIN | PLY | CGR |
| PEG | ICY | TRY | CMN |
| CUE | COB | TAU | PNT |
| ONE | GOT | PIU |
The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N, P, R, T. The number of words is 27, and these are all shown in the first three columns. The last word, PIU, is a musical term in common use; but although it has crept into some of our dictionaries, it is Italian, meaning " a little ; slightly." The remaining twenty-six are good words. Of course a TAU-cross is a T-shaped cross, also called the cross of St. Anthony, and borne on a badge in the Bishop's Palace at Exeter. It is also a name for the toad-fish.
We thus have twenty-six good words and one doubtful, obtained under the required conditions, and I do not think it will be easy to improve on this answer. Of course we are not bound by dictionaries but by common usage. If we went by the dictionary only in a case of this kind, we should find ourselves involved in prefixes, contractions, and such absurdities as I.O.U., which Nuttall actually gives as a word.
272.—THE NINE SCHOOLBOYS.
The boys can walk out as follows:—
| 1st Day. | 2nd Day. | 3rd Day. | ||||||
| A | B | C | B | F | H | F | A | G |
| D | E | F | E | I | A | I | D | B |
| G | H | I | C | G | D | H | C | E |
| 4th Day. | 5th Day. | 6th Day. | ||||||
| A | D | H | G | B | I | D | C | A |
| B | E | G | C | F | D | E | H | B |
| F | I | C | H | A | E | I | G | F |
Every boy will then have walked by the side of every other boy once and once only. Dealing with the problem generally, boys may walk out in triplets under the conditions on days, where may be nought or any integer. Every possible pair will occur once. Call the number of boys . Then every boy will pair times, of which times he will be in the middle of a triplet and times on the outside. Thus, if we refer to the solution above, we find that every boy is in the middle twice (making 4 pairs) and four times on the outside (making the remaining 4 pairs of his 8). The reader may now like to try his hand at solving the two next cases of 21 boys on 15 days, and 33 boys on 34 days. It is, perhaps, interesting to note that a school of 489 boys could thus walk out daily in one leap year, but it would take 731 girls (referred to in the solution to No. 269) to perform their particular feat by a daily walk in a year of 365 days.
273.—THE ROUND TABLE.
The history of this problem will be found in The Canterbury Puzzles (No. 90). Since the publication of that book in 1907, so far as I know, nobody has succeeded in solving the case for that unlucky number of persons, 13, seated at a table on 66 occasions. A solution is possible for any number of persons, and I have recorded schedules for every number up to 25 persons inclusive and for 33. But as I know a good many mathematicians are still considering the case of 13, I will not at this stage rob them of the pleasure of solving it by showing the answer. But I will now display the solutions for all the cases up to 12 persons inclusive. Some of these solutions are now published for the first time, and they may afford useful clues to investigators.
The solution for the case of 3 persons seated on 1 occasion needs no remark.
A solution for the case of 4 persons on 3 occasions is as follows:—
| 1 | 2 | 3 | 4 |
| 1 | 3 | 4 | 2 |
| 1 | 4 | 2 | 3 |
Each line represents the order for a sitting, and the person represented by the last number in a line must, of course, be regarded as sitting next to the first person in the same line, when placed at the round table.
The case of 5 persons on 6 occasions may be solved as follows:—
| 1 | 2 | 3 | 4 | 5 |
| 1 | 2 | 4 | 5 | 3 |
| 1 | 2 | 5 | 3 | 4 |
| 1 | 3 | 2 | 5 | 4 |
| 1 | 4 | 2 | 3 | 5 |
| 1 | 5 | 2 | 4 | 3 |
The case for 6 persons on 10 occasions is solved thus:—
| 1 | 2 | 3 | 6 | 4 | 5 |
| 1 | 3 | 4 | 2 | 5 | 6 |
| 1 | 4 | 5 | 3 | 6 | 2 |
| 1 | 6 | 2 | 5 | 3 | 4 |
| 1 | 2 | 4 | 5 | 6 | 3 |
| 1 | 3 | 5 | 6 | 2 | 4 |
| 1 | 4 | 6 | 2 | 3 | 5 |
| 1 | 5 | 2 | 3 | 4 | 6 |
| 1 | 6 | 3 | 4 | 5 | 2 |
It will now no longer be necessary to give the solutions in full, for reasons that I will explain. It will be seen in the examples above that the 1 (and, in the case of 5 persons, also the 2)
