Out of thin cardboard—old business cards answer this purpose nicely—make thirty-two blank counters, the size of a dime. Then upon a piece of note-paper mark off a figure just three inches square, and divide it by lines into nine compartments, each containing one square inch. The puzzle is, to arrange the counters in the external cells of the square four different times, and each time to have nine in a row, yet to have the sum of the counters different, and varying from twenty to thirty-two. If you will inspect the following figures you will see how this is possible: the first represents the original disposition of the counters in the cells of the square; the second, that of the same counters when four are taken away; the third, the manner in which they must be disposed when these four are brought back with four others; and the fourth with the addition of four more. There are always nine in each external row, and yet in the first case the whole number is twenty-four, in the second it is twenty, in the third twenty-eight, and in the fourth thirty-two. The numbers are substituted in the place of the counters in the above figures for convenience, but Fig. 5 represents the disposition of the counters, as indicated in Fig. 2.
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ANOTHER ARITHMETICAL TRICK.
By knowing the last figure of the product of any two numbers, to tell the other figures. If the number seventy-three be multiplied by each of the numbers in the following arithmetical progression, 3, 6, 9, 12, 15, 18, 21, 24, 27, the products will terminate with the nine digits, in this order, 9, 8, 7, 6, 5, 4, 3, 2, 1; the numbers themselves being as follows: 219, 438, 657, 876, 1095, 1314, 1533, 1752, and 1971. Let, therefore, a little bag be provided, consisting of two partitions, into one of which put several tickets, marked with the number 73, and into the other put as many tickets, 3, 6, 9, etc., up to 27. Then open that part of the bag containing the number 73, and ask a person to take out one ticket only; after which, dexterously change the opening, and desire another person to take a ticket from the other part. Let them now multiply their two numbers together, and tell you the last figure of the product, by which you will readily determine from the foregoing series what the remaining figures must be. Suppose, for example, the numbers taken out of the bag were 73 and 12, then as the product of these two numbers, which is 876, has 6 for its last figure, you will readily know it is the fourth of the series and the other two figures must be 8 and 7.
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TO TELL TWO OR MORE NUMBERS WHICH A PERSON HAS THOUGHT OF.
These numbers must not exceed 9. Let him think of two or three numbers, double the first and add 1 to the product, multiply the whole by 5, and add to that product the second number. If there be a third, make him double the first sum and add 1 to it; then desire him to multiple the new sum by 5, and to add to it the third number. If there should be a fourth number, you must proceed in the same manner, desiring him to double the preceding sum, to add 1 to it, to multiply by 5, and then 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 three, 55; if 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 of the rest.
Suppose the numbers thought of to be 3, 4, 6; by adding 1 to 6, the double of the first, we have 7, which being multiplied by 5 gives 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 by 5, the result will be 395. Lastly, if we add 6, the third number thought of, the sum will be 401, and if 55 be deducted from it we shall have for the remainder 346, the figures of which 3, 4, and 6, indicate in order the three numbers thought of.
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