131.—­THE SPANISH MISER.

There must have been 386 doubloons in one box, 8,450 in another, and 16,514 in the third, because 386 is the smallest number that can occur.  If I had asked for the smallest aggregate number of coins, the answer would have been 482, 3,362, and 6,242.  It will be found in either case that if the contents of any two of the three boxes be combined, they form a square number of coins.  It is a curious coincidence (nothing more, for it will not always happen) that in the first solution the digits of the three numbers add to 17 in every case, and in the second solution to 14.  It should be noted that the middle one of the three numbers will always be half a square.

132.—­THE NINE TREASURE BOXES.

Here is the answer that fulfils the conditions:—­

A = 4         B = 3,364      C = 6,724
D = 2,116     E = 5,476      F = 8,836
G = 9,409     H = 12,769     I = 16,129

Each of these is a square number, the roots, taken in alphabetical order, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the required difference between A and B, B and C, D and E. etc., is in every case 3,360.

133.—­THE FIVE BRIGANDS.

The sum of 200 doubloons might have been held by the five brigands in any one of 6,627 different ways.  Alfonso may have held any number from 1 to 11.  If he held 1 doubloon, there are 1,005 different ways of distributing the remainder; if he held 2, there are 985 ways; if 3, there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways; if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388 ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso held 11 doubloons, the remainder could be distributed in 3 different ways.  More than 11 doubloons he could not possibly have had.  It will scarcely be expected that I shall give all these 6,627 ways at length.  What I propose to do is to enable the reader, if he should feel so disposed, to write out all the answers where Alfonso has one and the same amount.  Let us take the cases where Alfonso has 6 doubloons, and see how we may obtain all the 704 different ways indicated above.  Here are two tables that will serve as keys to all these answers:—­

         Table I. Table II. 
    A = 6.  A = 6. 
    B = n.  B = n. 
    C = (63 — 5n) + m.  C = 1 + m. 
    D = (128 + 4n) — 4m.  D = (376 — 16n) — 4m. 
    E = 3 + 3m.  E = (15n — 183) + 3m.

In the first table we may substitute for n any whole number from 1 to 12 inclusive, and m may be nought or any whole number from 1 to (31 + n) inclusive.  In the second table n may have the value of any whole number from 13 to 23 inclusive, and m may be nought or any whole number from 1 to (93 — 4n) inclusive.  The first table thus gives (32 + n) answers for every value of n; and the second table gives (94 — 4n) answers for every value of n.  The former, therefore, produces 462 and the latter 242 answers, which together make 704, as already stated.