Problem. — To mark pictures, giving 3 x’s to 2 or 3, 2 to 4 or 5, and 1 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 in the 1st row, and in the 3rd; but only 7 in the end. Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the end.

Twenty-two answers have been received. Of these, 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong.

Of the eleven answers with which some working is supplied, 3 are wrong. C. H. begins with the rash assertion that under the given conditions “the sum is impossible. For”, he or she adds (these initialed correspondents are dismally vague beings to deal with: perhaps “it” would be a better pronoun), “10 is the least possible number of pictures&r............