## 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.

Read Answer
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THE ARTILLERYMEN'S DILEMMA.
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THE SULTAN'S ARMY.