to 9 or 10; also giving 3 o's to 1 or 2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the smallest possible number of pictures, and to give them the largest possible number of marks.
Answer.—10 pictures; 29 marks; arranged thus:—
| x | x | x | x | x | x | x | x | x | o |
|---|---|---|---|---|---|---|---|---|---|
| x | x | x | x | x | o | o | o | o | |
| x | x | o | o | o | o | o | o | o | o |
Solution.—By giving all the x's possible, putting into brackets the optional ones, we get 10 pictures marked thus:—
| x | x | x | x | x | x | x | x | x | (x) |
|---|---|---|---|---|---|---|---|---|---|
| x | x | x | x | (x) | |||||
| x | x | (x) |
By then assigning o's in the same way, beginning at the other end, we get 9 pictures marked thus:—
| (o) | o | |||||||
|---|---|---|---|---|---|---|---|---|
| (o) | o | o | o | |||||
| (o) | o | o | o | o | o | o | o | o |
All we have now to do is to run these two wedges as close together as they will go, so as to get the minimum number of pictures——erasing optional marks where by so doing we can run them closer, but otherwise letting them stand. There are 10 necessary marks