Of the more honourable, DELTA attempts a novelty--to discover _which_ son comes of age by elimination: it assumes, successively, that it is the middle one, and that it is the youngest; and in each case it _apparently_ brings out an absurdity. Still, as the proof contains the following bit of algebra, "63 = 7_x_ + 4_y_; [** therefore] 21 = _x_ + 4 sevenths of _y_," I trust it will admit that its proof is not _quite_ conclusive. The rest of its work is good. MAGPIE betrays the deplorable tendency of her tribe--to appropriate any stray conclusion she comes across, without having any _strict_ logical right to it. Assuming _A_, _B_, _C_, as the ages at first, and _D_ as the number of the years that have elapsed since then, she finds (rightly) the 3 equations, 2_A_ = _B_, _C_ = _B_ + _A_, _D_ = 2_B_. 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; [** therefore] 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 are 5 possible sets of ages, and eliminates four of them. Suppose that, instead of 5, there had been 5 million possible sets? Would SIMPLE SUSAN have courageously ordered in the necessary gallon of ink and ream of paper? The solution sent in by C. R. is, like that of SIMPLE SUSAN, partly tentative, and so does not rise higher than being Clumsily Right.