272. THE NINE SCHOOLBOYS. This is a new and interesting companion puzzle to the "Fifteen Schoolgirls" (see solution of No. 269), and even in the simplest possible form in which I present it there are unquestionable difficulties. Nine schoolboys walk out in triplets on the six week days so that no boy ever walks _side by side_ with any other boy more than once. How would you arrange them?

If we represent them by the first nine letters of the alphabet, they (1/2)

Follow

SOLUTION TO 272. THE NINE SCHOOLBOYS. (1/3)

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, 12n+9 boys may walk out in triplets

SOLUTION TO 272. THE NINE SCHOOLBOYS. (3/3)

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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 24 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.

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Dudeney's Amusements@dudeney_puzzles@mathstodon.xyzSOLUTION TO 272. THE NINE SCHOOLBOYS. (2/3)

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under the conditions on 9n+6 days, where n may be nought or any integer. Every possible pair will occur once. Call the number of boys m. Then every boy will pair m-1 times, of which (m-1)/4 times he will be in the middle of a triplet and (m-1)/2 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