1897 - 98 .] Prof. Tait on Generalization of Josephus' Problem. 167 
Here the numbers, and their order, are the same, but those in 
the lower rank are three places in advance.- 
By fives 
12 10 351 11 8742 9 60 
1 2 345 6 789 10 11 12 
5 3 10 7 1 13 11 4 6 2 12 9 8 
1 2 3 4 5 6 7 8 9 10 11 12 13 
The numbers of the first line, increased by units, and those of 
the third, are 
13 11 4 6 2 12 9 8 5 3 10 7 1 
5 3 10 7 1 13 11 4 6 2 12 9 8 , 
again the same order, hut now shifted forwards by five places. 
It is easy to see that the two rows thus formed are identical 
when m = n + 1 . Thus 
By tens : — 
1428637950 
123456789 
2539748 10 6 1 
1234567 8 9 10 
and the statement above is obviously verified. 
To show how rapidly the results of this process can be extended 
to higher numbers, I confine myself to the Josephus question, as 
regards himself alone, the last man. For the others, the mode of 
procedure is exactly the same. 
Given that the final survivor in 41, told off by threes, is the 
31st, we have 
n last man. 
41 31 
The rule just given shows that succeeding numbers in these 
columns are formed as follows : — taking only those which com- 
mence, as it were, a new cycle : — 
41 +x 31 + 3x -(41+x) = 2x- 10. 
The value of x which makes the right hand side one or other of 
1, and 2, is therefore to be chosen, so we must put a; = 6, and the 
result is 
47 
q 
