+ 5 squared= 15; and so on. So if we want to form a
triangle with 8 counters on each side we shall require half of 8 +
8 squared, or 36 counters. This is a pretty little property of numbers.
Before going further, I will here say that if the reader refers to the
"Stonemason's Problem" (No. 135) he will remember that the sum of any
number of consecutive cubes beginning with 1 is always a square, and
these form the series 1 squared, 3 squared, 6 squared, 10 squared, etc. It will now be understood
when I say that one of the keys to the puzzle was the fact that these
are always the squares of triangular numbers--that is, the squares of 1,
3, 6, 10, 15, 21, 28, etc., any of which numbers we have seen will form
a triangle.

Every whole number is either triangular, or the sum of two triangular
numbers or the sum of three triangular numbers. That is, if we take any
number we choose we can always form one, two, or three triangles with
them. The number 1 will obviously, and uniquely, only form one triangle;
some numbers will only form two triangles (as 2, 4, 11, etc.); some
numbers will only form three triangles (as 5, 8, 14, etc.). Then, again,
some numbers will form both one and two triangles (as 6), others both
one and three triangles (as 3 and 10), others both two and three
triangles (as 7 and 9), while some numbers (like 21) will form one, two,
or three triangles, as we desire. Now for a little puzzle in triangular
numbers.

Sandy McAllister, of Aberdeen, practised strict domestic economy, and
was anxious to train his good wife in his own habits of thrift. He told
her last New Year's Eve that when she had saved so many sovereigns that
she could lay them all out on the table so as to form a perfect square,
or a perfect triangle, or two triangles, or three triangles, just as he
might choose to ask he would add five pounds to her treasure. Soon she
went to her husband with a little bag of L36 in sovereigns and claimed
her reward. It will be found that the thirty-six coins will form a
square (with side 6), that they will form a single triangle (with side
8), that they will form two triangles (with sides 5 and 6), and that
they will form three triangles (with sides 3, 5, and 5). In each of the
four cases all the thirty-six coins are used, as required, and Sandy
therefore made his wife the promised present like an honest man.

The Scotsman then undertook to extend his promise for five more years,
so that if next year the increas