the figures of every number, of which it is a multiple, forms a multiple of 9.

Thus, the product of 17 multiplied by 9 is 153, and the sum of these figures 1 + 5 + 3 = 9. In the like manner, the sum of the figures 6777, which is the product of 753 multiplied by 9, is equal to 27, a multiple of 9.

This will not appear astonishing when we reflect, that twice 9 is equal to 18, three times 927, &c.; where it is evident that the tens and units of the product are always reciprocal complements of 9.

If we take any two numbers whatever, one of them, or their sum, or their difference, will always be divisible by three. This may be so easily proved, that it is needless to illustrate it by examples.

Every square number must necessarily terminate with two ciphers, or by 1, 4, 5, 6, 9; and this may enable us to determine, at one view, whether a numerical quantity be a square or not. (10.)

The number 2, of all the whole numbers, is the only one the sum and product of which are equal. Thus 2 24, as well as 2 x 2. But in fractional numbers we find other two quantities, the sum and product of which are in like manner equal.

For this purpose, the sum of the two numbers must be divided by each of them separately. The two fractions which thence arise, will give the same quantity, both when added and multiplied. (5. 7.)

If we take the two numbers 2 and 5, and divide their sum by each of them separately, the two fractions 7 ,, will give the same result when added, as well as when multiplied. This may be easily proved by any person in the least acquainted with vulgar fractions. (2.)

Every square number (10.) is divisible by 3, or becomes so when diminished by unity. This may be easily proved with any square number whatever. Thus, 4 — 1, 16 — 1, 25 — 1, 49 — 1, 121 — 1, &c. are all divisible by 3; and the case is the same with the rest. Every square number is divisible by 4, or becomes. so when diminished by unity.

Every square number is divisible also by 5, or becomes so when increased or diminished by unity.

Every odd square, diminished by unity, is a mul tiple of 8.

Every power of 5 terminates with 5, and every power of 6 with 6. (10.)

PROBLEM.

To find Two Numbers, the Squares of which, if added together, shall form a Square Number.

If any two numbers whatever be multiplied together, the double of their product will be one of the two numbers sought; and the difference of their squares will be the other.

Thus, if the numbers 2 and 3, the squares of which are 4 and 9, be multiplied together, their product will be 6; if we then take 12, the double of this product, and 5 the difference of their squares, we shall have two numbers, the sum of the squares of which will be a square number; for their squares are 144 and 25, which by addition give 169, the square of 13.

OF ARITHMETICAL AND GEOMETRICAL PROGRESSION, WITH SOME PROBLEMS WHICH DEPEND ON THEM.

SECTION I.

Arithmetical Progression, with an Explanation of its Principal Properties.

Any series of numbers, continually increasing or decreasing by the same quantity, forms what is called an arithmetical progression. (19.)

Thus, the series of numbers 1, 2, 3, 4, 5, 6, &c. or I, 5, 9, 13, &c. or 20, 18, 16, 14, 12, &c. or 15, 12, 9, 6, 3, are arithmetical progressions; for, in the first, the difference between each term and the followng one, which exceeds it, is always 1; in the second

it is 4; it is 2 also in the third, which goes on decreasing, and 3 in the fourth.

From this definition of arithmetical progression, the following consequences may be deduced:

1st. Any term of an arithmetical progression, is equal to the first, plus the common difference taken as many times as there are terms before it. (22.)

2nd. The sum of the extremes is always equal to the sum of any two terms equally distant from them; or double the mean term, if the progression contains an odd number of terms. (21.)

3rd. If the sum of the extremes be multiplied by half the number of terms, when the terms are even, or the mean by the whole number of terms when the latter are odd, the product will be the sum of the progression.

By considering, with a little attention, the following progressions, the truth of these consequences will be readily perceived.

14 17.

9.

If a hundred stones are placed in a straight line, at the distance of a yard from each other, the first being at the same distance from a basket; how many yards must the person walk, who engages to pick them up, one by one, and to put them into the basket?

It is evident that, to pick up the first stone, and put it into the basket, the person must walk two yards; for the second he must walk 4; for the third 6; and so on increasing by two, to the hundredth.

The number of yards, therefore, which the person must walk, will be equal to the sum of the progression 2, 4, 6, &c. the last term of which is 200. (22.) But the sum of the progression is equal to 202, the sum of the two extremes, multiplied by 50, or half the number of terms; that is to say, 10,100 yards, which makes more than 5 miles.

PROBLEM II.

A gentleman employed a bricklayer to sink a well, and agreed to give him at the rate of three shillings for the first yard in depth, 5 for the second, 7 for the third, and so on, increasing to the twentieth, where he expected to find water: how much was due to the bricklayer when he had completed the work?

This question may be easily answered by the rules before given, for the difference of the terms is 2, and the number of terms 20; consequently, to find the twentieth term, we must multiply 2 by 19, and add 38, the product, to the first term 3, which will give for the twentieth term 41. (22.)

If we then add the first and last terms, that is to say, 3 and 41, which will make 44, and multiply this sum by 10, or half the number of terms, the product 440 will be the sum of all the terms of the progression, or the number of shillings due to the bricklayer, when he completed the work. (21.)

He would therefore have to receive 221.

PROBLEM III.

A merchant being considerably in debt, one of his creditors, to whom he owed £1860. offered to give him an acquittance, on condition of his agreeing to pay the whole sum in twelve monthly installments; that is to say, £100. the first month, and to increase the payment by a certain sum each succeeding month to the twelfth inclusive, when the whole debt would be discharged. By what sum was the payment of each month increased?

In this problem we have given the first term 100, the number of the terms 12, and their sum 1860; but the common difference of the terms is unknown.

This difference may be found in the following manner: As the sum of the extremes, in an even arithmetical progression, is equal to the sum total, divided

by half the number of terms, if the sum total 1860 be divided by 6, or half the number of terms, we shall have 310 for the sum of the first and last term, from which if we subtract 100, the first term, the remainder 210 will be the last term; but the last term is always equal to the first and the common difference taken as many times as there are terms before it. If we, therefore, deduct the first term 100, from 210 the last, and divide 110, the remainder, by 11, we shall have 10 as the required difference. The first term being 100, the second therefore will be 110, the third 120, &c. (21.)

PROBLEM IV.

A gentleman employed a bricklayer to sink a well, to the depth of 20 yards, and agreed to give him £20. for the whole; but the bricklayer happening to die when he had completed only 8 yards, how much was due to his heirs?

To imagine that two fifths of the whole sum were due to the workman, because 8 yards are two fifths of the whole depth, would be erroneous; for as the difficulty must increase arithmetically as the depth, it is natural to suppose that the price should increase in the same ratio.

To resolve this problem, therefore, £20. or 400 shillings, must be divided into twenty terms in arithmetical progression; and the sum of the first eight of these will express what was due to the bricklayer for his labour.

But 400 shillings may be divided into twenty terms in arithmetical progression a great many different ways, according to the value of the first term, which is here undetermined: if we suppose it, for example, to be 1 shilling, the progression will be 1, 3, 5, 7, &c. the last term of which will be 39; and consequently the sum of the first eight terms will be 64 shillings.

But to resolve the problem in a proper manner, so as to give to the bricklayer his just due for the com