A and B have an 8 gallon cask full of wine, which they wish to divide into two equal parts, and the only measures they have are a 5-gallon cask and a 3-gallon cask. How shall they make the division with these two vessels?

Method.—First fill the 3-gallon cask from the 8-gallon cask; then pour these 3 gallons into the 5-gallon cask; then fill the 3-gallon cask again, and fill the 5-gallon cask from the 3-gallon cask; this will leave 1 gallon in the 3-gallon cask; then empty the 5-gallon cask into the 8-gallon cask, pour the 1 gallon from the 3-gallon cask into the 5-gallon cask, and fill the 3-gallon cask from the 8-gallon cask. There will then be 4 gallons in the 8-gallon cask.

Two men in an oyster saloon laid a wager as to which could eat the most oysters. One ate ninety-nine and the other ate a hundred, and won. How many did both eat?

Remark.—The catch is on a hundred and won. When spoken it sounds as if it meant “one ate ninety-nine and the other ate a hundred and one;” hence, the result usually given is two hundred. The correct result is one hundred and ninety-nine.

If a room with 8 corners had a cat in each corner, seven cats before each cat, and a cat on each cat’s tail, what would be the whole number of cats?

Answer.—Eight cats.

Tell a person to think of a number, multiply by 3, multiply the product by 2, divide the result by 6, add 20, subtract the number thought of, divide by 4, and then tell him what his result is.

Method.—The result will be five. The reason is clear. By multiplying by 3 and 2 and dividing by 6 he has obtained the number thought of. Add 20, he has the number thought of, plus 20; then subtract the number thought of, and he has twenty. Now I know he has twenty; hence, I can tell him what he has if he divides by 4.

A farmer having an ox-chain consisting of 15 links, broke it into five equal parts, and took it to a blacksmith to be welded together. The blacksmith agreed to repair it for 50 cents for each welding; but when he presented his bill he charged for four weldings, making the bill $2.00. The farmer objected to the bill, saying that it should have been repaired with only three weldings. How was it to be done?

Method.—Each piece consisted of three links; cut open the three links of one piece and use these to connect the other four pieces of the chain.

Think of a number, multiply by six, divide by three, add forty, divide by two; name the result, and I will name the number thought of.

Method.—Multiplying by six and dividing by three gives twice the number; add forty we have twice the number, plus forty, divide by two we have once the number, plus twenty; hence, if I subtract twenty from the result he gives me I have the number thought of.

Let a person select a number greater than 1 and not exceeding 10. I will add to it a number not exceeding 10, alternately with himself; and, although he has the advantage in selecting the number to start with, I will reach the even hundred first.

Method.—I make my additions so that the sums are 12, 23, 34, 45, etc., to 89, when it is evident I can reach the hundred first. With one who does not know the method, I need not run through the entire series, but merely aim for 89, and when the secret of this is seen aim at 78, then 67, etc.

Think of a number of 3 or more figures, divide by nine, and name the remainder; erase one figure of the number, divide by 9, and tell me the remainder and I will tell you what figure you erased.

Method.—If the second remainder is less than the first, the figure erased is the difference between the remainders; but if the second remainder is greater than the first, the figure erased equals 9, minus the difference of the remainders.

Let a person think of any number on the dial face of a watch. I will then point to various numbers, and at each he will silently add one to the number selected, until he arrives at twenty, which he will announce aloud; and my pointer will then be on the number he selected.

Method.—I point promiscuously about the face of the watch until the eighth point, which should be on the “12.” I then pass regularly around toward the “1” pointing at “11,” “10,” “9,” etc., until “twenty” is called, when my pointer will be over the number selected.

Take nine from six and ten from nine and fifty from forty, and six will remain.

Two-thirds of six is nine, one-half of twelve is seven, the half of five is four, and six is half of eleven.

Method.—Two-thirds of SIX is IX, the upper half of XII is VII, the half of FIVE is IV, and the upper half of XI is VI.

Two men have 24 ounces of fluid which they wish to divide between them equally. How shall they effect the division, provided they have only three vessels; one containing 5 oz., the other 11 oz., and the third 13 oz.?

Method.—The method is similar to the division of 8 gallons in the question on page 78.

Three persons own 51 quarts of rice, and have only two measures; one a 4-quart, the other a 7-quart measure. How shall they divide it into three equal parts?

Method.—One-third of 51 is 17; so each must have 17 quarts. To measure 17 quarts fill the 7-quart measure twice and pour into some large vessel, making 14 quarts; then fill the 7-quart measure, draw off 4 quarts in the 4-quart measure, and then pour the remaining 3 quarts in the vessel containing the 14 quarts.

Think of a number composed of two unequal digits, invert the digits, take the difference between this and the original number, name one of the digits and I will name the other.

Method.—The sum of the digits in the difference is always nine; hence, when one is named the other equals 9, minus the one named.

Take any number, consisting of three consecutive digits and permutate them, making 6 numbers, and take the sum of these numbers, divide by 6, and tell me the result and I will tell you the digits of the number taken.

Method.—The quotient consists of three equal digits; the digits of the number taken are: 1st. one of these equal digits; 2d. this digit increased by a unit; 3d. this digit diminished by a unit. The same principle holds when the digits of the number taken differ by 2, 3, or 4. It is a very pretty problem to prove that the sum is always divisible by 9 and 18.

Take any number, divide it by 9, and name the remainder. Multiply the number by some number which I name, and divide the product by 9, and I will name the remainder.

Method.—To tell the remainder, I multiply the first remainder by the number by which I told them to multiply the given number, and divide this product by 9. The remainder is the second number obtained.

Think of a number greater than 3, multiply it by 3; if even, divide it by 2; if odd, add 1, and then divide by 2. Multiply the quotient by 3; if even, divide by 2; if odd, add 1, and then divide by 2. Now divide by 9 and tell the quotient without the remainder, and I will tell you the number thought of.

Method.—If even both times, multiply the quotient by 4; if even 2d and odd 1st, multiply by 4 and add 1; if even 1st and odd 2d, multiply by 4 and add 2; if odd both times, multiply by 4 and add 3.

Suppose it were possible for a man in Cincinnati to start on Sunday noon, when the sun is in the meridian, and travel westward with the sun so that it might be in his meridian all the time. Now it was Sunday noon when he started, it has been noon with him all the way round, and is Monday noon when he returns. The question is, at what point did it change from Sunday noon to Monday noon?

Take any number, subtract the sum of the digits, strike out any digit from the remainder, tell me the sum of the remaining digits, and I will tell you the digit struck out.

Method.—Subtract the “sum of the remaining digits” from the smallest multiple of nine greater than “the sum.” The remainder will be the digit struck out.

In the bottom of a well, 45 feet deep, there was a frog which commenced traveling toward the top. In his journey he ascended 3 feet every day, but fell back 2 feet every night. In how many days did he get out?

Method.—He gains 1 foot a day, and in 42 days he is 3 feet from the top; and on the 43d day he reaches the top.

Think of any three numbers less than 10. Multiply the first by 2 and add 5 to the product. Multiply this sum by 5 and add the second number to the product. Multiply the last result by 10 and add the third number to the product; then subtract 250. Name the remainder and I will name the numbers thought of and in the order in which they were thought of.

Method.—The three digits composing this remainder will be the numbers thought of: and the order in which they were thought of will be the order of hundreds, tens, and units.

If a man had a triangular lot of land, the largest side being 136 rods, and each of the other sides 68 rods; what would be the value of the grass on it at the rate of $10 an acre?

Remark.—The “catch” in this is that the sides given will form no triangle.

Says A to B: “Give me four weights and I can weigh any number of pounds not exceeding 40.” Required the weights and the method of weighing.

Answer.—The weights are 1, 3, 9, and 27 pounds. In weighing we must put one or more in both scales, or some in one scale and some in another: thus, 7 lbs. = 9 lbs. + 1 lb. - 3 lbs.

Three men traveling with their wives came to a river which they wished to cross. There was but one boat and but two could cross at one time; and since the husbands were jealous no woman could be with a man unless her own husband was present. In what manner did they get across the river?

Method.—Let the persons be denoted A, B, and C, and Mrs. A, Mrs. B, and Mrs. C. First Mr. A and Mrs. A go over; then A comes back and Mrs. B and Mrs. C go over; then Mrs. A comes back and Mr. B and Mr. C go over; then Mr. B and Mrs. B return and Mr. A and Mr. B go over; then Mrs. C returns and Mrs. A and Mrs. B go over; then Mr. C returns and takes his wife, Mrs. C, over.

A man having a fox, a goose, and some corn came to a river which it was necessary to cross. He could, however, take only one across at a time, and if he left the goose and corn while he took the fox over, the goose would eat the corn; but if he left the fox and goose, the fox would kill the goose. How shall he get them all safely over?

Method.—First he takes the goose over, then returns and takes the fox over, then brings the goose back and takes the corn over, and then returns and takes the goose over again.

How may the 9 digits be arranged in a rectangular form so that the sum of any row, whether horizontal, vertical, or diagonal, shall equal 15?

Answer.—As below.

How may the first 16 digits be arranged so that the sum of the vertical, the horizontal, and the two oblique rows may equal 34?

Answer.—As below.

In what manner may the first 25 digits be arranged so that the sum of each row of five figures may equal 65?

Answer.—As below.

An old Jew took a diamond cross to a jeweler to have the diamonds reset, and fearing the jeweler might be dishonest, he counted the diamonds and found that they numbered 7 in three different ways. Now the jeweler stole two diamonds, but arranged the remainder so that they counted 7 each way as before. How was it done?

Method.—The form of the cross when left is represented by Fig. 1, and when returned by Fig. 2. It will be seen by the figures how the diamonds were counted by the old Jew, and how they were arranged by the jeweler, who “jewed” the Jew.

Take 10 pieces of money, lay them in a row, and require some one to put them together into heaps of two in each heap by passing each piece over two others.

Method.—Let the pieces be denoted by the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Then place 7 on 10, 5 on 2, 3 on 8, 1 on 4, and 9 on 6.

A man goes to a store and purchases a pair of boots worth $5 and hands out a $50 bill to pay for them. The merchant, not being able to make the change, goes over the street to a broker and gets the bill changed and then returns and gives the man who bought the boots his change. After the purchaser of the boots had been gone a few hours the broker, finding the bill to be a counterfeit, comes and demands $50 of good money from the merchant. How much does the merchant lose?

Remark.—At first glance some say $45 and the boots, some $50 and the boots, some $95 and the boots, and others $100 and the boots. Which is correct?

A vessel with a crew of 30 men, half of whom were black, became short of provisions and fearing that unless half the crew were thrown overboard all would perish, the captain proposed to the sailors to stand upon deck in a row and every ninth man be thrown overboard until half the crew were destroyed. It so happened that the whites were saved. Required, the order of arrangement.

Answer.—W W W W B B B B B W W B W W W B W B B W W B B B W B B W W B. This can easily be proved by trial, using letters or figures to represent men.

Suppose a hare is 10 rods before a hound, and that the hound runs 10 rods while the hare runs 1 rod. Now, when the hound has run 10 rods the hare has run 1 rod; hence they are now 1 rod apart, and when the hound has run that one rod the hare has run 1/10 of a rod; hence they are now 1/10 of a rod apart, and when the hound has run the 1/10 of a rod they are 1/100 of a rod apart; and in the same way it may be shown the hare is always 1/10 of the previous distance ahead of the hound; hence the hound can never catch the hare. How is the contrary shown mathematically? How far will the hound run to catch the hare.

Answer.—The distance the hound runs will be represented by the series

to infinity. The sum of this series can be found by the algebraic formula

in which a = 10 and r = 1/10. Substituting the value of a and r we have

This may be solved more simply as follows: The hound runs 10 times as fast as the fox, hence 10 times the distance the fox runs equals the distance the hound runs. Then 10 times the distance the fox runs, minus once the distance the fox runs, which is 9 times the distance the fox runs, is 10 rods; and once the distance the fox runs is 1/9 of 10 rods, or 10/9 rods; and 10 times the distance the fox runs, or the distance the hound runs, is 10 times 10/9 or 100/9, or 11-1/9 rods.

If through passenger trains, running to and from Philadelphia and San Francisco daily, start at the same hour from each place (difference of longitude not being considered) and take the same time—seven days—for the trip, how many through trains will the Pacific Express, that leaves the San Francisco depot at 9 P. M. Sunday, have met when it reaches the Philadelphia depot?

Answer.—As the Pacific Express starts from San Francisco, a train which left Philadelphia the previous Sunday reaches San Francisco, which is not to be counted as a meeting of trains. There are, however, six other trains on the way which it will meet. Also, a train starts from Philadelphia on the same Sunday as the train starts from San Francisco, another on Monday, another on Tuesday, etc., up to Saturday—that is, seven trains, all of which it meets, making, with the six trains previously started, thirteen trains in all which it meets. A train leaves Philadelphia on Sunday at the same time the Pacific Express reaches there, but this is not counted as a meeting.

A switch siding to a single-track railroad is just long enough to clear a train of eight cars and a locomotive. How can two trains of sixteen cars and a locomotive, each going in opposite directions, pass each other at this siding and each locomotive remain with, and have the same relative position to its own train after as before passing?

Answer.—Let one train and its locomotive be denoted by A, and the other train and locomotive by B, and let the track be denoted by a b and the siding by c d, and suppose train A to be going in the direction of a b, and train B in the direction of b a. Then let locomotive B, with eight cars, run out toward a, past c, and back up on the siding with its eight cars; then let train A run out toward b, past c; then let B draw its eight cars on to the main track and run out toward a; then let train A back over toward a, past c, and locomotive A be detached from train A and run over toward b and connect with the eight cars of train B and draw them over past c, and back them up on the siding, and then run off the siding and connect again with its own cars and run on toward b, past c; then let locomotive B back its eight cars and, turning on the siding, connect the two halves of its train and move off past a, the train A moving on at the same time past b.

A and B went to market with 30 pigs each. A sold his pigs at 2 for $1, and B sold his pigs at the rate of 3 for $1, and they, together, received $25. The next day A went to market alone with 60 pigs, and, wishing to sell at the same rate, sold them 5 for $2, and received only $24. Why should he not receive as much as when B owned half of the pigs?

Answer.—The rate of 2 pigs for $1 is 1 pig for $½, and the rate of 3 pigs for $1 is 1 pig for $?; the average rate is 2 pigs for $½ + $?, or $?, or 1 pig for $5/12. The rate of 5 pigs for $2 is 1 pig for $?. So it is seen that the reason A did not receive as much is that he sold his pigs at a less rate than when they both went to market.

Two hunters killed a deer and sold it by the pound in the woods. They had no proper means of weighing it, but knew their own weights—one 130 pounds and the other 190 pounds. They placed a rail across a fence so that it balanced with one of them on each end. They then exchanged places, the lighter man taking the deer in his lap, and the rail again balanced; what was the weight of the deer?

Answer.—Let the weight of the deer be denoted by D; then, by the principles of the lever, we have the proportion:

Who can solve the following problem?

Explanation.—CI, CLI, CLIO (Clio, the muse of history, one of the nine muses).

Suppose the figure to represent railroad tracks, C D and E F being each the length of a car or locomotive, and a and b each representing a car on the track and c representing a locomotive on E F. Now how can the locomotive change the relative position of a and b so that b will be on the track where a is and a will be on the track where b is?

Answer.—The locomotive c backs a down and out toward A, then runs over toward B and backs b up on E F, then runs back toward B and goes over toward A, then runs up C E and draws b down on C E, then runs over toward A and gets car a, draws it over toward B and backs it up on D E. It can also be readily done by first backing a down on A C and drawing to and leaving it on C D.

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