s, to the
other three, least?"

_Answer._--"From No. 9."

* * * * *

[Illustration]

Let A be No. 9, B No. 25, C No. 52, and D No. 73.

Then AB = [** sqrt](12^{2} + 5^{2}) = [** sqrt]169 = 13;
AC = 21;
AD = [** sqrt](9^{2} + 8^{2}) = [** sqrt]145 = 12 +
(N.B. _i.e._ "between 12 and 13.")
BC = [** sqrt](16^{2} + 12^{2}) = [** sqrt]400 = 20;
BD = [** sqrt](3^{2} + 21^{2}) = [** sqrt]450 = 21+;
CD = [** sqrt](9^{2} + 13^{2}) = [** sqrt]250 = 15+;

Hence sum of distances from A is between 46 and 47; from B, between 54
and 55; from C, between 56 and 57; from D, between 48 and 51. (Why not
"between 48 and 49"? Make this out for yourselves.) Hence the sum is
least for A.

* * * * *

Twenty-five solutions have been received. Of these, 15 must be marked
"0," 5 are partly right, and 5 right. Of the 15, I may dismiss
ALPHABETICAL PHANTOM, BOG-OAK, DINAH MITE, FIFEE, GALANTHUS NIVALIS
MAJOR (I fear the cold spring has blighted our SNOWDROP), GUY, H.M.S.
PINAFORE, JANET, and VALENTINE with the simple remark that they insist
on the unfortunate lodgers _keeping to the pavement_. (I used the words
"crossed to Number Seventy-three" for the special purpose of showing
that _short cuts_ were possible.) SEA-BREEZE does the same, and adds
that "the result would be the same" even if they crossed the Square, but
gives no proof of this. M. M. draws a diagram, and says that No. 9 is
the house, "as the diagram shows." I cannot see _how_ it does so. OLD
CAT assumes that the house _must_ be No. 9 or No. 73. She does not
explain how she estimates the distances. Bee's Arithmetic is faulty: she
makes [** sqrt]169 + [** sqrt]442 + [** sqrt]130 = 741. (I suppose you
mean [** sqrt]741, which would be a little nearer the truth. But roots
cannot be added in this manner. Do you think [** sqrt]9 + [** sqrt]16 is
25, or even [** sqrt]25?) But AYR'S state is more perilous still: she
draws illogical conclusions with a frightful calmness. After pointing
out (rightly) that AC is less than BD she says, "therefore the nearest
house to the other three must be A or C." And again, after pointing out
(rightly) that B and D are both within the half-square containing A,
she says "therefore" AB + AD must be less than BC + CD. (There is no
logical force in either "therefore." For the first, try Nos. 1, 21, 60,
70: this will make your premiss true, and your conclusion false.
Si