, 8, 11, 14, 17-Arithmetical Progression. 5, 15, 45, 135, 405—Geometrical Progression.

In the case of arithmetical progression, as above or in any other manner exemplified, it may be noticed that the amount of the first and last term is always the same as twice the amount of the middle term; thus 5 and 17 being 22, are equal to twice II, or 22. The cause of this is, that as the numbers increase or decrease in equal degrees, the last number is just as much more as the first is less than the number in the middle; and the two being added, the amount must consequently be double the central number. The same rule holds good with respect to any two numbers at equal distances from the number in the middle. If the series be an even number, and do not possess a middle term, then the two terms nearest the middle (called the mean terms) must be added together: thus in the natural series from 1 to 24, 12 and 13 are the two nearest the middle, and one being added to the other makes 25. the sum of the first and last term.

In geometrical progression, each term is a factor of all the numbers or terms that follow, and a product of all that go before, so that there is an harmonious ratio pervading the whole. Each term bears an exact proportion to its predecessor, because the multiplier is the same. Supposing, as above, the multiplier to be 3, the term 15 is proportionally greater than 5, as 45 is greater than 15. In the technical language of arithmetic, as 15 is to 5, so is 45 to 15. To save words such a proposition is written down with dots, thus-15:5:45: 15. The two dots mean is to, and four dots mean so is. The same formula is applicable to any series of proportional terms, though not in continued proportion to each other.

In order to discover the ratio between any two terms we divide the largest by the least, and the quotient is the ratio : 45 divided by 15 gives 3 as the ratio. By thus ascertaining the ratio of two terms, we are furnished with the means of arriving at the ratio of other terms. We cannot do better than explain the method of working out this principle in the ratio of numbers, by giving the following passages from the admirable Lessons on Arithmetic, by Mr. T. Smith of Liverpool. Taking the four regularly advancing terms, 15, 45, 405, and 1215, he proceeds: 'Suppose that we had only the first three, and that it were our wish to find the fourth, which term bears the same proportion to the third as the second does to the first. The thing we have first to do, is to discover the ratio between the first and second terms, in order to do which, as before shown, we divide the larger by the smaller, and this gives us the ratio 3, with which, by multiplying the third term, we produce the fourth; or, let the three terms be these, 405, 1215, 5, and let it be our wish to find a fourth which shall bear the same relation to the 15 as 1215 does to 405. divide and multiply as before, and the fourth term is produced. And in this manner, having two numbers, or two quantities of any kind, bearing a certain proportion towards each other, and a third, to which we would find a number or quantity that should bear a like proportion, in this manner do we proceed, and thus easily may we find the number we require."

We

Referring to the discovered ratio of 45 to 15 to be 3, or the fifteenth part-" Now" (continues this author), "what would

have been the consequence had we multiplied the third term (405) by the whole, instead of by a fifteenth part of the second? The consequence would have been, that we should have had a term or number fifteen times larger than that required. But this would be a matter of no difficulty; for it would be set right at once and our purpose gained, by dividing the overlarge product by 15. Let us write this process down : 405 × 45 18225, and 18225 ÷ 15 = 1215,-which 1215 bears the same proportion to 405 as does 45 to 15. And this is the rule, when the terms are properly placed—multiplying the second and third terms together, and dividing the product by the first; this avoids all difficulties arising from the occurrence of fractions in the course of the process, and gives us, in all cases, any proportional terms we may require."

Rule of Three.

N the principle now explained, we can, in any affairs of business, ascertain the amount of an unknown quantity, by knowing the amount of other three quantities, which, with the unknown quantity, bear a proportional relation. The word quantity is here used, but any sum of money is also

meant.

Let it be remembered, that the ratio of one number to another is the number of times that the former contains the latter; for example, the ratio of 6 to 3 is 2, that of 12 to 4 is 3, and that of 8 to 12 is 3. When two numbers have the same ratio as other two, they constitute a proportion. Thus, the ratio of 8 to 6 is the same as that of 12 to 9, and the equality of these two ratios is represented thus:

8:6 12:9, or, 8:6:: 12:9.

The following is the rule for stating and working questions:Make that term which is of the same kind as the answer sought, the second or middle term, Consider, from the nature of the question, whether the answer should be more or less than this term; if more, make the smaller of the other two terms the first, and the greater the third; if the answer should be less than the middle term, make the greater of the two terms the first, and the smaller the third; then multiply the second and third terms together and divide the result by the first term. The quotient found will be the answer to the question, and it will be found to bear the same proportion to the third term as the second does to the first.

Such is the principle of working Rule of Three questions, whatever be their apparent complexity. If either the first or third term, or both, include fractional parts, they must be reduced to the denomination of the fractions before working; thus if one be reduced to shillings, the other must be made shillings also; if to pence, both must be pence, and so

An easy and uniform method of computing interest, D. Fish's method, is to place the principal, the rate, and the time in months, on the right of a vertical line, and 12 on the left; or, if the time is short and contains days, reduce to days, and place 360 on the left. After canceling equal factors on both sides of the line, the product of the remaining factors on the right, divided by the factor, if any, on the left, will give the required interest.

To find the interest of $184.80 for 1 yr. 5 mo. at 5%.

per

cent.

the rate %, we

have $12.25,

6

6.125 Int. for 6 mo.

the int. for I

2

1.020 Int. for I mo.

yr. Then, for

.510 Int. for 15 da.

6 mo. take

of 1 year's int.,

$19.905= Int. for 1 yr. 7 mo. 15 da. for I mo. take

TO TELL ANY NUMBER THOUGHT OF.

ESIRE any person to think of a number, say a certain number of shillings; tell him to borrow that sum of some one in the company, and add the number borrowed to the amount thought of. It will here be proper to name the person who lends him the shillings and to beg the one who makes the calculation to do it with great care, as he may readily fall into an error, especially the first time. Then, say to the person-"I do not lend you, but give you To, add them to the former sum." Continue in this manner -"Give the half to the poor, and retain in your memory the other half." Then add :-" Return to the gentleman, or lady, what you borrowed, and remember that the sum lent you was exactly equal to the number thought of." Ask the person if he knows exactly what remains. He will answer "Yes." You must then say-" And I know, also, the number that remains; it is equal to what I am going to conceal in my hand." Put into one of your hands five pieces of money, and desire the person to tell how many you have got. He will answer five; upon which open your hand, and show him the five pieces. You may then say "I well knew that your result was five, but if you had thought of a very large number, for example, two or three millions, the result would have been much greater, but my hand would not have held a number of pieces equal to the remainder." The person then supposing that the result of the calculation must be different, according to the difference of the number thought of, will imagine that it is necessary to know the last number in order to guess the result : but this idea is faise; for, in the case which we have here supposed, whatever be the number thought of, the remainder must always be five. The reason of this is as follows:-The sum, the half of which is given to the poor, is nothing else than twice the number thought of, plus 10; and when the poor have received their part, there remains only the number thought of, plus 5; but the number thought of is cut off when the sum borrowed is returned, and, consequently, there remain only 5

It may be hence seen that the result may be easily known, since it will be the half of the number given in the third part of the operation; for example, whatever be the number thought of, the remainder will be 36 or 25, according as 72 or

50 have been given. If this trick be performed several times successively, the number given in the third part of the operation must be always different; for if the result were several times the same, the deception might be discovered. When the first five parts of the calculation for obtaining a result are finished, it will be best not to name it at first, but to continue the operation, to render it more complex, by saying, for example :"Double the remainder, deduct 2, add 3, take the fourth part," etc.; and the different steps of the calculation may be kept in mind, in order to know how much the first result has been increased or diminished. This irregular process never fails to confound those who attempt to follow it.

A Second Method.-Bid the person take I from the number thought of, and then double the remainder; desire him to take I from the double, and to add to it the number thought of; in the last place, ask him the number arising from this addition, and, if you add 3 to it, the third of the sum will be the number thought of. The application of this rule is so easy, that it is needless to illustrate it by an example.

A Third Method.-Desire the person to add I to the triple of the number thought of, and to multiply the sum by 3; then bid him add to this product the number thought of. and the result will be a sum, from which, if 3 be subtracted, the remainder will be ten times the number required; and if the cipher on the right be cut off from the remainder, the other figure will indicate the number sought.

Example:-Let the number thought of be 6, the triple of which is 18; and if I be added, it makes 19; the triple of this last number is 57, and if 6 be added, it makes 63, from which, if 3 be subtracted, the remainder will be 60; now, if the cipher on the right be cut off, the remaining figure, 6, will be the number required.

A Fourth Method.-Bid the person multiply the number thought of by itself; then desire him to add I to the number thought of, and to multiply it also by itself; in the last place. ask him to tell the difference of these two products, which will certainly be an odd number, and the least half of it will be the number required. Let the number thought of, for example, be 10, which, multiplied by itself, give 100; in the next place, 10 increased by I is II, which, multiplied by itself. makes 121; and the difference of these two squares is 21, the least half of which, being 10, is the number thought of. This operation might be varied by desiring the person to multiply the second number by itself, after it has been diminished by 1.

in this case, the number thought of will be equal to the greater half of the difference of the two squares. Thus, in the preceding example, the square of the number thought of is 100, and that of the same number less 1, is 81; the difference of these is 19; the greater half of which, or 10, is the number thought of.

TO TELL TWO OR MORE NUMBERS
THOUGHT OF.

If one or more of the numbers thought of be greater than 9, we must distinguish two cases; that in which the number or the numbers thought of is odd, and that in which it is even. In the first case, ask the sum of the first and second, of the second and third, the third and fourth, and so on to the last, and then the sum of the first and the last. Having written

down all these sums in order, add together all those, the places of which are odd, as the first, the third, the fifth, etc.; make another sum of all those, the places of which are even, as the second, the fourth, the sixth, etc., subtract this sum from the former, and the remainder will be the double of the first number. Let us suppose, for example, that the five following numbers are thought of, 3, 7, 13, 17, 20, which, when added two and two as above, give 10, 20, 30, 37, 23: the sum of the first, third, and fifth, is 63, and that of the second and fourth is 57; if 57 be subtracted from 63, the remainder, 6, will be the double of the first number, 3, Now, if 3 be taken from 10, the first of the sums, the remainder, 7, will be the second number, and by proceeding in this manner we may find all the rest.

In the second case, that is to say, if the number or the numbers thought of be even, you must ask and write down, as above, the sum of the first and the second, that of the second and third, and so on, as before; but, instead of the sum of the first and last, you must take that of the second and last; then add together those which stand in the even places, and form them into a new sum apart; add also those in the odd places, the first excepted, and subtract this sum from the former, the remainder will be the double of the second number; and if the second number, thus found, be subtracted from the sum of the first and second, you will have the first number; if it be taken from that of the second and third, it will give the third; and so of the rest. Let the numbers thought of be, for example, 3, 7, 13, 17; the sums formed as above are 10, 20, 30, 24, the sum of the second and fourth is 44, from which, if 30, the third, be subtracted, the remainder will be 14, the double of 7, the second number. The first, therefore, is 3, the third 13, and the fourth 17.

When each of the numbers thought of does not exceed 9, they may be easily found in the following manner :

Having made the person add 1 to the double of the first number thought of, desire him to multiply the whole by 5, and to add to the product the second number. If there be a third, make him double this first sum, and add 1 to it; after which, desire him to multiply the new sum by 5, and to add to it the third number. If there be a fourth, proceed in the same manner, desiring him to double the preceding sum, to add to it I, to multiply by 5, to add the fourth number, and

so on.

Then ask the number arising from the addition of the last number thought of, and if there were two numbers, subtract 5 from it; if there were three, 55; if there were four, 555, and so on, for the remainder will be composed of figures, of which the first on the left will be the first number thought of, the next the second, and so on.

Suppose the number thought of to be 3, 4, 6; by adding I to 6, the double of the first, we shall have 7, which, being multiplied by 5, will give 35; if 4, the second number thought of, be then added, we shall have 39, which, doubled, gives 78; and, if we add 1, and multiply 79, the sum, by 5, the result will be 395. In the last place, if we add 6, the number thought of, the sum will be 401; and if 55 be deducted from it, we shall have, for remainder, 346, the figures of which, 3. 4, 6, indicate in order the three numbers thought of.

THE MONEY GAME.

A person having in one hand a piece of gold, and in the other a piece of silver, you may tell in which hand he has the gold, and in which the silver, by the following method:-Some value, represented by an even number, such as 8, must be assigned to the gold; and a value represented by an odd number, such as 3, must be assigned to the silver; after which, desire the person to multiply the number in the right hand, by any even number whatever, such as 2; and that in the left hand by an odd number, as 3; then bid him add together the two products, and if the whole sum be odd, the gold will be in the right hand, and the silver in the left: if the sum be even, the contrary will be the case.

To conceal the artifice better, it will be sufficient to ask whether the sum of the two products can be halved without a remainder; for in that case the total will be even, and in the contrary case odd.

It may be readily seen, that the pieces, instead of being in the two hands of the same person, may be supposed to be in the hands of two persons, one of whom has the even number, or piece of gold, and the other the odd number, or piece of silver. The same operations may then be performed in regard to these two persons, as are performed in regard to the two hands of the same persons, calling the one privately the right, and the other the left.

THE GAME OF THE RING.

This game is an application of one of the methods employed to tell several numbers thought of, and ought to be performed in a company not exceeding nine, in order that it may be less complex. Desire any one of the company to take a ring, and put it on any joint of whatever finger he may think proper. The question then is, to tell what person has the ring, and on what hand, what finger, and on what joint.

For this purpose, you must call the first person 1, the second 2, the third 3, and so on. You must also denote the ten fingers of the two hands by the following numbers of the natural progression, I, 2, 3, 4, 5, etc., beginning at the thumb of the right hand, and ending at that of the left, that this order of the number of the finger may, at the same time, indicate the hand. In the last place, the joints must be denoted by I, 2. 3, beginning at the points of the fingers.

To render the solution of this problem more explicit, let us suppose that the fourth person in the company has the ring on the sixth finger, that is to say, on the little finger of the left hand, and on the second joint of that finger.

Desire some one to double the number expressing the person, which, in this case, will give 8; bid him add 6 to this double, and multiply the sum by 5, which will make 65; then tell him to add to this product the number denoting the finger, that is to say 6, by which means you will have 71; and, in the last place, desire him to multiply the last number by 10, and to add to the product the number of the joint, 2; the last result will be 712; if from this number you deduct 250, the remainder will be 462; the first figure of which, on the left, will denote the person; the next, the finger, and, consequently, the hand; and the last, the joint.

It must here be observed, that when the last result contains a cipher, which would have happened in the present example had the number of the figure been 10, you must privately subtract from the figure preceding the cipher, and assign the value of 10 to the cipher itself.

THE GAME OF THE BAG.

To let a person select several numbers out of a bag, and to tell him the number which shall exactly divide the sum of those he had chosen :-Provide a small bag, divided into two parts, into one of which put several tickets, numbered 6, 9, 15, 36, 63, 120, 213, 309, etc., and in the other part put as many other tickets, marked No. 3 only. Draw a handful of tickets from the first part, and after showing them to the company, put them into the bag again, and having opened it a second time, desire any one to take out as many tickets as he thinks proper; when he has done that, you open privately the other part of the bag, and tell him to take out of it one ticket only. You may safely pronounce that the ticket shall contain the number by which the amount of the other numbers is divisible; for, as each of these numbers can be multiplied by 3, their sum total must, evidently, be divisible by that number. An ingenious mind may easily diversify this exercise, by marking the tickets in one part of the bag with any numbers that are divisible by 9 only, the properties of both and 3 being the same; and it should never be exhibited to the same company twice without being varied.

THE CERTAIN GAME.

Two persons agree to take, alternately, numbers less than a given number, for example, II, and to add them together till one of them has reached a certain sum, such as 100. By what means can one of them infallibly attain to that number before the other?

The whole artifice in this consists in immediately making choice of the numbers 1, 12, 23, 34, and so on, or of a series which continually increases by II, up to 100. Let us suppose that the first person, who knows the game, makes choice of I; it is evident that his adversary, as he must count less than II, can at most reach II, by adding 10 to it. The first will then take I, which will make 12; and whatever number the second may add, the first will certainly win, provided he continually

add the number which forms the complement of that of his adversary to II; that is to say, if the latter take 8, he must take 3: if 9, he must take 2; and so on. By following this method he will infallibly attain to 89: and it will then be impossible for the second to prevent him from getting first to 100; for whatever number the second takes he can attain only to 99; after which the first may say "and I makes 100." If the second take 1 after 89, it would make 90, and his ad. versary would finish by saying-"and 10 make 100." Between two persons who are equally acquainted with the game, he who begins must necessarily win.

If your opponent have no knowledge of numbers, you may take any other number first, under 10, provided you subsequently take care to secure one of the last terms, 56, 67, 78, etc., or you may even let him begin, if you take care afterward to secure one of these numbers.

This exercise may be performed with other numbers; but, in order to succeed, you must divide the number to be attained by a number which is a unit greater than what you can take each time, and the remainder will then be the number you must first take. Suppose, for example, the number to be attained be 52, and that you are never to add more than 6; then, dividing 52 by 7, the remainder, which is 3, will be the number which you must first take; and whenever your opponent adds a number you must add as much to it as will make it equal to 7, the number by which you divided, and so in continua. tion.

ODD OR EVEN.

Every odd number multiplied by an odd number produces an odd number; every odd number multiplied by an even number produces an even number; and every even number multiplied by an even number also produces an even number. So, again, an even number added to an even number, and an odd number added to an odd number, produce an even number; while an odd and even number added together produce an odd number.

If any one holds an odd number of counters in one hand, and an even number in the other, it is not difficult to discover in which hand the odd or even number is. Desire the party to multiply the number in the right hand by an even number, and that in the left hand by an odd number, then to add the two sums together, and tell you the last figure of the product; if it is even, the odd number will be in the right hand; and if odd, in the left hand; thus, supposing there are 5 counters in the right hand, and 4 in the left hand, multiply 5 by 2, and 4 by 3, thus:-5 x 2 = 10, 4 × 3 12, and then adding IC to 12, you have 10 +12= 22, the last figure of which, 2, is even, and the odd number will consequently be in the right hand.

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