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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