Part 9

SOLUTION.--9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. -------------------------------------- 8 + 6 + 4 + 1 + 9 + 7 + 5 + 3 + 2 = 45.

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.

76542 -24 24 ----- 76518

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?

86,214 42,680 57,319 62,854 37,145 ------ 286,212

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.

67,856 47,218 52,781 ------ 167,855

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.

Think of 7 Double it 14 Add 10 to it 10 -- Halve it 2 ) 24 -- Which will leave 12 Subtract the number thought of 7 -- The remainder will be 5

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.