A STUDY IN THRIFT.
(
Money Puzzles)
Certain numbers are called triangular, because if they are taken to
represent counters or coins they may be laid out on the table so as to
form triangles. The number 1 is always regarded as triangular, just as 1
is a square and a cube number. Place one counter on the table--that is,
the first triangular number. Now place two more counters beneath it, and
you have a triangle of three counters; therefore 3 is triangular. Next
place a row of three more counters, and you have a triangle of six
counters; therefore 6 is triangular. We see that every row of counters
that we add, containing just one more counter than the row above it,
makes a larger triangle.
Now, half the sum of any number and its square is always a triangular
number. Thus half of 2 + 2 squared = 3; half of 3 + 3 squared = 6; half of 4 +
4 squared = 10; half of 5 + 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 increased number of sovereigns that she has
saved can be laid out in the same four different ways she will receive a
second present; if she succeeds in the following year she will get a
third present, and so on until she has earned six presents in all. Now,
how many sovereigns must she put together before she can win the sixth
present?
What you have to do is to find five numbers, the smallest possible,
higher than 36, that can be displayed in the four ways--to form a
square, to form a triangle, to form two triangles, and to form three
triangles. The highest of your five numbers will be your answer.
Answer:
Mrs. Sandy McAllister will have to save a tremendous sum out of her
housekeeping allowance if she is to win that sixth present that her
canny husband promised her. And the allowance must be a very liberal one
if it is to admit of such savings. The problem required that we should
find five numbers higher than 36 the units of which may be displayed so
as to form a square, a triangle, two triangles, and three triangles,
using the complete number in every one of the four cases.
Every triangular number is such that if we multiply it by 8 and add 1
the result is an odd square number. For example, multiply 1, 3, 6, 10,
15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which are
the squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every case
where 8x squared + 1 = a square number, x squared is also a triangular. This point
is dealt with in our puzzle, "The Battle of Hastings." I will now merely
show again how, when the first solution is found, the others may be
discovered without any difficulty. First of all, here are the figures:--
8 x 1 squared + 1 = 3 squared
8 x 6 squared + 1 = 17 squared
8 x 35 squared + 1 = 99 squared
8 x 204 squared + 1 = 577 squared
8 x 1189 squared + 1 = 3363 squared
8 x 6930 squared + 1 = 19601 squared
8 x 40391 squared + 1 = 114243 squared
The successive pairs of numbers are found in this way:--
(1 x 3) + (3 x 1) = 6 (8 x 1) + (3 x 3) = 17
(1 x 17) + (3 x 6) = 35 (8 x 6) + (3 x 17) = 99
(1 x 99) + (3 x 35) = 204 (8 x 35) + (3 x 99) = 577
and so on. Look for the numbers in the table above, and the method will
explain itself.
Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and
1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and
40391; and they will also form single triangles with sides of 8, 49,
288, 1681, 9800, and 57121. These numbers may be obtained from the last
column in the first table above in this way: simply divide the numbers
by 2 and reject the remainder. Thus the integral halves of 17, 99, and
577 are 8, 49, and 288.
All the numbers we have found will form either two or three triangles at
will. The following little diagram will show you graphically at a glance
that every square number must necessarily be the sum of two triangulars,
and that the side of one triangle will be the same as the side of the
corresponding square, while the other will be just 1 less.
[Illustration
+-----------+
+---------+ |. . . . ./.|
|. . . ./.| |. . . ./. .|
|. . ./. .| |. . ./. . .|
|. ./. . .| |. ./. . . .|
|./. . . .| |./. . . . .|
/. . . . .| /. . . . . .|
+---------+ +-----------+
]
Thus a square may always be divided easily into two triangles, and the
sum of two consecutive triangulars will always make a square. In numbers
it is equally clear, for if we examine the first triangulars--1, 3, 6,
10, 15, 21, 28--we find that by adding all the consecutive pairs in turn
we get the series of square numbers--9, 16, 25, 36, 49, etc.
The method of forming three triangles from our numbers is equally
direct, and not at all a matter of trial. But I must content myself with
giving actual figures, and just stating that every triangular higher
than 6 will form three triangulars. I give the sides of the triangles,
and readers will know from my remarks when stating the puzzle how to
find from these sides the number of counters or coins in each, and so
check the results if they so wish.
+----------------------+-----------+---------------+-----------------------+
| Number | Side of | Side of | Sides of Two | Sides of Three |
| | Square. | Triangle. | Triangles. | Triangles. |
+------------+---------+-----------+---------------+-----------------------+
| 36 | 6 | 8 | 6 + 5 | 5 + 5 + 3 |
| 1225 | 35 | 49 | 36 + 34 | 33 + 32 + 16 |
| 41616 | 204 | 288 | 204 + 203 | 192 + 192 + 95 |
| 1413721 | 1189 | 1681 | 1189 + 1188 | 1121 + 1120 + 560 |
| 48024900 | 6930 | 9800 | 6930 + 6929 | 6533 + 6533 + 3267 |
| 1631432881 | 40391 | 57121 | 40391 + 40390 | 38081 + 38080 + 19040 |
+------------+---------+-----------+---------------+-----------------------+
I should perhaps explain that the arrangements given in the last two
columns are not the only ways of forming two and three triangles. There
are others, but one set of figures will fully serve our purpose. We thus
see that before Mrs. McAllister can claim her sixth L5 present she must
save the respectable sum of L1,631,432,881.
