EXAMPLE. Suppose I should fix upon the number 50, and ask you to tell which I thought on; you would ask me in which of the columns, and how many, does it stand in? I would say, in the second and two last columns. You would then put together 2+16+32 =50, and tell me to a certainty; because 50 stands no where else but in the second and two last columns. Whatever number you guess upon, if only in the first, second, third, fourth, fifth, or sixth columns, it must be 1, 2, 4, 8, 16, or 32; because each of them are found but once; but where it takes two or more of the figures in the horizontal row of geometricals, to make the one you think on, the number is to be found in all the columns, exactly under the geometricals which compose it, and no where else. Hence it is easy both to make and understand them; and even to extend the cards to much higher numbers. If another column was added, its head number would be 64; from which a table could be generated, that would give all the numbers from 1 to 127; because 63+64-127. 32. Divide 40 into four such parts, that if to the first you add 4, from the second subtract 4, multiply the third by 4, and divide the fourth by 4, the sum, difference, product, and quotient will be all equal. 9. Ans.,,, and . 33. Into what three whole parts can 117 be divided, so that if each of them be squared, their squares will be in arithmetical ratio? Ans. 9, 45, 63. 34. Divide 50 into four such parts, that when each of those parts are cubed, the sum of their cubes may be a cube. COMPUTATION OF SHOT. Cannon-balls and shells are usually piled up in a pyramidal, or prismatic form; the base being either an equilateral triangle, a square, or a rectangle. In the triangle and square, the pile is a pyramid, finished by a single ball at the top; but, when the base is a rectangle, the pile is finished by a single row of balls. In pyramids of balls, the number of balls on each side of the base decreases from the base upwards by 1; that is, if the number of balls in one side of the base be 20, the number in the tier next above that will be 19, and in the next 18, &c. until all the sides of the pyramid meet at the top in a single ball. Therefore, if the base be a triangle, the several tiers of balls will be triangular numbers, which, reckoned from the top, are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, &c. But, if the base be square, the tiers of balls will be square numbers, which, beginning at the top, are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, &c. 1. To find the number of shot in a triangular pile. Count the number of balls in one side of the base; to the number add 1, and to the same number add 2; then multiply these three numbers, one by the other, and the product by the third, and divide the product by 6; the quotient is the number of shot in the pile. EXAMPLE. How many balls are there in a finished triangular pile, one side of the base consisting of 24 balls? By adding 1 and 2 to 24, the three numbers are 24 x 25 x 26 24, 25, 26, and 2600 balls. Ans. 6 2. To find the number of shot in a square pyramidal pile. Count the number of balls in one side of the base; to the number add 1, and to its double add 1; then multiply these three numbers into one another continually, and divide the product by 6; the quotient is the number of balls in the pile. EXAMPLE. It is required to find the number of shot in a square pile, each side of its base consisting of 20 balls. By adding 1 to 20, and 1 to the double of 20, the three numbers are 20, 21, 41, and 20×21×41 2870 balls. Ans. 6 In rectangular piles, each horizontal tier is a rectangle, the uppermost being one row of balls; and the number of horizontal tiers is equal to the number in the breadth of the base. 3. To find the number of shot in an oblong pile. Compute the number of shot in a square pile, each side of its base being the breadth of the rectangular base. Then take the difference between the length and breadth of the rectangular base; multiply the breadth + 1 by half of the breadth, and multiply the product by the said difference, the last product is the number of balls in the prismatic pile; and the sum of the two is the number of balls in the whole pile. EXAMPLE. There is a complete oblong pile of 15 tiers, the number of balls in the top row being 32; it is required to find the number of shot in this pile. Here, the difference between the length and breadth of the base is 31, and the breadth of the base is 15. A square pile, the side of whose base is 15, is =1240. The breadth of the base + 1 16 x 31 x 15 6 is 16, and the half breadth is 7, and 16 × 71⁄2 × 31= 3720. Lastly, 1240+3720-4960. Ans. 4. To find the number of shot in a broken oblong pile. To twice the length, and to twice the breadth of the uppermost tier, add the number of tiers - 1, and multiply the two sums together; also multiply the 1, and number of tiers + 1, by the same number add one third of this product to the former; then one fourth part of the sum, multiplied by the number of tiers, shall give the number of shot in the oblong broken pile. EXAMPLE. Suppose there is a broken oblong pile of shot, the length of the uppermost tier being 25, and its breadth 16 balls, and the number of tiers 11; it is required to find the number of shot in the pile. Here twice the length is 50, and twice the breadth .1 is 10. is 32, and the number of tiers Then 60 x 42=2520, is the first product, also 12 × 10=120 is the second product, one third of which is 40, and 2520+40=2560. Lastly, 2560 X 11 4 |