How to become a scientist

As the principal object of these articles is to enable the young reader to learn something in his sports, and to understand what he is doing, we shall, before proceeding to the curious tricks and feats connected with the science of numbers, present him with some arithmetical aphorisms, upon which most of the following examples are founded:

Aphorisms of Number.

1. If two even numbers be added together, or subtracted from each other, their sum or difference will be an even number.

2. If two uneven numbers be added or subtracted, their sum or difference will be an even number.

3. The sum or difference of an even and an uneven number added or subtracted will be an uneven number.

4. The product of two even numbers will be an even number, and the product of two uneven numbers will be an uneven number.

5. The product of an even and uneven number will be an even number.

6. If two different numbers be divisible by any one number, their sum and their difference will also be divisible by that number.

7. If several different numbers, divisible by 3, be added or multiplied together, their sum and their product will also be divisible by 3.

8. If two numbers divisible by 9, be added together, their sum of the figures in the amount will be either 9 or a number divisible by 9.

9. If any number be multiplied by 9, or by any other number divisible by 9, the amount of the figures of the product will be either 9 or a number divisible by 9.

10. In every arithmetical progression, if the first and last term be each multiplied by the number of terms, and the sum of the two products be divided by 2, the quotient will be the sum of the series.

11. In every geometric progression, if any two terms be multiplied together, their product will be equal to that term which answers to the sum of these two indices. Thus, in the series:

If the third and fourth terms, 8 and 16, be multiplied together, the product, 128, will be the seventh term of the series. In like manner, if the fifth term be multiplied into itself, the product will be the tenth term; and if that sum be multiplied into itself, the product will be the twentieth term. Therefore, to find the last, or twentieth term of a geometric series, it is not necessary to continue the series beyond a few of the first terms.

Previous to the numerical recreations, we shall here describe certain mechanical methods of performing arithmetical calculations, such as are not only in themselves entertaining, but will be found more or less useful to the young reader.

To Find a Number Thought of.

FIRST METHOD.

Let him inform you what is the number produced; it will always end with 3. Strike off the 3, and inform him that he thought of 6.

SECOND METHOD.

Let him inform you what is the number produced. You must then, in every case, subtract 320; the remainder is, in this example, 600; strike off the 2 ciphers, and announce 6 as the number thought of.

THIRD METHOD.

Desire a person to think of a number—say 6. He must then proceed:

Which will be the number thought of.

FOURTH METHOD.

Desire a person to think of a number—say 6. He must then proceed as follows:

Which you can say is the number he thought of.

FIFTH METHOD.

You can then tell him that if he will subtract from this the number he thought of, the remainder will be, in the case supposed, 2.

Note.—The remainder is always half the number you tell him to add.

To Discover Two or More Numbers that a Person has Thought of.

FIRST CASE.

Where each of the numbers is less than 10. Suppose the numbers thought of were 2, 3, 5.

And to proceed in the same manner for as many numbers as were thought of. Let him tell you the last sum produced (in this case, 290). Then, if there were two numbers thought of, you must subtract 5; if three, 55; if four, 555. You must here subtract 55; leaving a remainder of 235, which are the numbers thought of, 2, 3, and 5.

SECOND CASE.

Where one or more of the numbers are 10, or more than 10, and where there is an odd number of numbers thought of.

Suppose he fixes upon five numbers, viz., 4, 6, 9, 15, 16.

He must add together the numbers as follows, and tell you the various sums:

You must then add together the 1st, 3d, and 5th sums, viz., 10 + 24 + 20 = 54, and the 2d and 4th, 15 + 31 = 46; take one from the other, leaving 8. The half of this is the first number, 4; if you take this from the sum of the 1st and 2d you will have the 2d number, 6; this taken from the sum of the 2d and 3d will give you the 3d, 9; and so on for the other numbers.

THIRD CASE.

Where one or more of the numbers are 10, or more than 10, and where an even number of numbers has been thought of.

Suppose he fixes on six numbers, viz: 2, 6, 7, 15, 16, 18. He must add together the numbers as follows, and tell you the sum in each case:

You must then add together the 2d, 4th, and 6th sums, 13 + 31 + 24 = 68, and the 3d and 5th sums, 22 + 34 = 56. Subtract one from the other, leaving 12; the 2d number will be 6, the half of this; take the 2d from the sum of the 1st and 2d, and you will get the 1st; take the 2d from the sum of the 2d and 3d, and you will have the 3d, and so on.

How Many Counters Have I in My Hands?

A person having an equal number of counters in each hand, it is required to find how many he has altogether.

Suppose he has 16 counters, or 8 in each hand. Desire him to transfer from one hand to the other a certain number of them, and to tell you the number so transferred. Suppose it be 4, the hands now contain 4 and 12. Ask him how many times the smaller number is contained in the larger; in this case it is three times. You must then multiply the number transferred, 4, by the 3, making 12, and add the 4, making 16; then divide 16 by the 3 minus 1; this will bring 8, the number in each hand.

In most cases fractions will occur in the process; when 10 counters are in each hand and if four be transferred, the hands will contain 6 and 14.

He will divide 14 by 6 and inform you that the quotient is 2 1/3.

You multiply 4 by 2 1/3, which is 9 1/3.

Add four to this, making 13 1/3 equal to 40/3.

Subtract 1 from 2 1/3, leaving 1 1/3 or 4/3.

Divide 40/3 by 4/3, giving 10, the number in each hand.

The Three Travelers.

Three men met at a caravansary or inn, in Persia; and two of them brought their provisions along with them, according to the custom of the country; but the third, not having provided any, proposed to the others that they should eat together, and he would pay the value of his proportion. This being agreed to, A produced 5 loaves, and B 3 loaves, all of which the travelers ate together, and C paid 8 pieces of money as the value of his share, with which the others were satisfied, but quarreled about the division of it. Upon this the matter was referred to the judge, who decided impartially. What was his decision?

At first sight it would seem that the money should be divided according to the bread furnished; but we must consider that as the 3 ate 8 loaves, each one ate 2 2/3 loaves of the bread he furnished. This from 5 would leave 2 1/3 loaves furnished the stranger by A; and 3 - 2 2/3 = 1/3 furnished by B, hence 2 1/3 to 1/3 = 7 to 1, is the ratio in which the money is to be divided. If you imagine A and B to furnish, and C to consume all, then the division will be according to amounts furnished.

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 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 person, calling the one privately the right, and the other the left.

The Philosopher’s Pupils.

To find a number of which the half, fourth, and seventh, added to three, shall be equal to itself.

This was a favorite problem among the ancient Grecian arithmeticians, who stated the question in the following manner: “Tell us, illustrious Pythagoras, how many pupils frequent thy school?” “One-half,” replied the philosopher, “study mathematics, one-fourth natural philosophy, one-seventh preserve silence, and there are three females besides.”

The answer is 28: 14 + 7 + 4 + 3 = 28.

The Certain Game.

Two persons agree to take, alternately, numbers less than a given figure, for example, 11, 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 11, up to 100. Let us suppose that the first person, who knows the game, makes choice of 1, it is evident that his adversary, as he must count less than 11, can at most reach 11, by adding 10 to it. The first will then take 1, 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 11; 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 1 makes 100.” If the second take 1 after 89, it would make 90, and his adversary would finish by saying—“and 10 make 100.” Between two persons who are equally acquainted with the game, he who begins must necessarily win.

The Dice Guessed Unseen.

A pair of dice being thrown, to find the number of points on each die without seeing them. Tell the person who cast the dice to double the number of points upon one of them, and add 5 to it; then to multiply the sum produced by 5, and to add to the product the number of points upon the other die. This being done, desire him to tell you the amount, and having thrown out 25, the remainder will be a number consisting of two figures, the first of which, to the left, is the number of points on the first die, and the second figure, to the right, the number of the other. Thus:

Suppose the number of points of the first die which comes up to be 2, and that of the other 3; then, if to 4, the double of the points of the first, there be added 5, and the sum produced, 9, be multiplied by 5, the product will be 45; to which, if 3, the number of points on the other die, be added, 48 will be produced, from which, if 25 be subtracted, 23 will remain; the first figure of which is 2, the number of points on the first die, and the second figure 3, the number on the other.

The Famous Forty-five.

How can number 45 be divided into four such parts that, if to the first part you add 2, from the second part you subtract 2, the third part you multiply by 2, and the fourth part you divide by 2, the sum of the addition, the remainder of the subtraction, the product of the multiplication, and the quotient of the division, be all equal?

Required to subtract 45 from 45, and leave 45 as a remainder.

The Astonished Farmer.

A and B took each 30 pigs to market. A sold his at 3 for a dollar, B at 2 for a dollar, and together they received $25. A afterwards took 60 alone, which he sold as before, at 5 for $2, and received but $24: what became of the other dollar?

This is rather a catch question, the insinuation that the first lot were sold at the rate of 5 for $2, being only true in part. They commence selling at that rate, but after making ten sales, A’s pigs are exhausted, and they have received $20; B still has 10, which he sells at “two for a dollar,” and of course receives $5; whereas had he sold them at the rate of 5 for $2, he would have received but $4. Hence the difficulty is easily settled.

The Expunged Figure.

In the first place we desire a person to write down secretly, in a line, any number of figures he may choose, and add them together as units; having done this, tell him to subtract that sum from the line of figures originally set down; then desire him to strike out any figure he pleases, and add the remaining figures in the line together as units (as in the first instance), and inform you of the result, when you will tell him the figure he has struck out.

Suppose, for example, the figures put down are 76542; these added together, as units, make a total of 24; deduct twenty-four from the first line, and 76518 remain; if 5, the center figure, be struck out, the total will be 22. If 8, the first figure, be struck out, 19 will be the total.

In order to ascertain which figure has been struck out, you make a mental sum one multiple of 9 higher than the total given. If 22 be given as the total, then 3 times 9 are 27, and 22 from 27 show that 5 was struck. If 19 be given, that sum deducted from 27 shows 8.

Should the total be equal multiples of 9, as 18, 27, 36, then 9 has been expunged.

With very little practice, any person may perform this with rapidity: it is therefore needless to give any further examples. The only way in which a person can fail in solving this riddle is when either a number 9 or a 0 is struck out, as it then becomes impossible to tell which of the two it is, the sum of the figures in the line being an even number of nines in both cases.

Mysterious Addition.

It is required to name the quotient of five or three lines of figures—each line consisting of five or more figures—only seeing the first line before the other lines are even put down. Any person may write down the first line of figures for you. How do you find the quotient?

When the first line of figures is set down, subtract 2 from the last right-hand figure, and place it before the first figure of the line, and that is the quotient for five lines. For example, suppose the figures are 86,214, the quotient will be 286,212. You may allow any person to put down the two first and the fourth lines, but you must always set down the third and fifth lines, and in doing so always make up 9 with the line above.

Therefore in the annexed diagram you will see that you have made 9 in the third and fifth lines with the lines above them. If the person you request to put down the figures should set down a 1 or 0 for the last figure, you must say: “We will have another figure,” and another, and so on until he sets down something above 1 or 2.

In solving the puzzle with 3 lines, you subtract 1 from the last figure, and place it before the first figure, and make up the third line yourself to 9. For example: 67,856 is given, and the quotient will be 167,855, as shown in the above diagram.

The Remainder.

A very pleasing way to arrive at an arithmetical sum, without the use of either slate or pencil, is to ask a person to think of a figure, then to double it, then add a certain figure to it, now halve the whole sum, and finally to abstract from that the figure first thought of. You are then to tell the thinker what is the remainder.

The key to this lock of figures is, that half of whatever sum you request to be added during the working of the sum is the remainder. In the example given, 5 is the half of 10, the number requested to be added. Any amount may be added, but the operation is simplified by giving only even numbers, as they will divide without fractions.

The Three Jealous Husbands.

Three jealous husbands, A, B and C, with their wives being ready to pass by night over a river, find at the water-side a boat which can carry but two at a time, and for want of a waterman they are compelled to row themselves over the river at several times. The question is, how those six persons shall pass, two at a time, so that none of the three wives may be found in the company of one or two men, unless her husband be present?

This may be effected in two or three ways; the following may be as good as any: Let A and wife go over—let A return—let B’s and C’s wives go over—A’s wife returns—B and C go over—B and wife return, A and B go over—C’s wife returns, and A’s and B’s wives go over—then C comes back for his wife. Simple as this question may appear, it is found in the works of Alcuin, who flourished a thousand years ago, hundreds of years before the art of printing was invented.

The Arithmetical Mouse-Trap.

One of the best and most simple mouse-traps in use may be constructed as follows: Get a slip of smooth pine, about the eighth of an inch thick, a quarter of an inch broad, and of sufficient length to cut out the following parts of a trap: First, an upright piece, three or four inches high, which must be square at the bottom, and a small piece to be cut off at the top to fit a notch in No. 2.

The second piece must be of the same length as the first, with the notch cut across nearly at the top of it, to fit the top of No. 1, and the other end of it trimmed to catch the notch in No. 3. The third piece should be twice as long as either of the others; a notch, similar to that in No. 2, must be cut in one end of it to catch the lower end of No. 2. Having proceeded thus far, you must put the pieces together, in order to finish it, by adding another notch in No. 3, the exact situation of which you will discover as follows: Place No. 1 upright, then put the notch of No. 2 in the thinned top of No. 1; then get a flat piece of wood, or a slate, one end of which must rest on the ground, and the center of the edge of the other on the top of No. 2. You will now find the thinned end of No. 2 elevated by the weight of the flat piece of wood or slate; then put the thinned end of it in the notch of No. 3, and draw No. 2 down by it, until the whole forms a resemblance of a figure 4; at the exact place where No. 3 touches the upright, cut a notch, which, by catching the end of No. 1, will keep the trap together. You may now bait the end of No. 3 with pieces of cheese; a mouse, by nibbling the bait, will pull down No. 3, the other pieces immediately separate, and the slate or board falls upon the mouse. We have seen numbers of mice, rats and birds caught by this.