APPENDIX.

She then says "supposing that A = 1, then B = 2, C = 3, and D = 4. Therefore for A , B , C , D , four numbers are wanted which shall be to each other as 1:2:3:4." It is in the "therefore" that I detect the unconscientiousness of this bird. The conclusion is true, but this is only because the equations are "homogeneous" ( i.e. having one "unknown" in each term), a fact which I strongly suspect had not been grasped—I beg pardon, clawed—by her. Were I to lay this little pitfall, " A + 1 = B , B + 1 = C ; supposing A = 1, then B = 2 and C = 3. Therefore for A , B , C , three numbers are wanted which shall be to one another as 1:2:3," would you not flutter down into it, oh Magpie, as amiably as a Dove? Simple Susan is anything but simple to me . After ascertaining that the 3 ages at first are as 3:2:1, she says "then, as two-thirds of their sum, added to one of them, = 21, the sum cannot exceed 30, and consequently the highest cannot exceed 15." I suppose her (mental) argument is something like this:—"two-thirds of sum, + one age, = 21; ∴ sum, + 3 halves of one age, = 31 and a half. But 3 halves of one age cannot be less than 1 and-a-half (here I perceive that Simple Susan would on no account present a guinea to a new-born baby!) hence the sum cannot exceed 30." This is ingenious, but her proof, after that, is (as she candidly admits) "clumsy and roundabout." She finds that there