Page:Methods of Operating the Comptometer (1895).djvu/38

 greater than the amount represented by the red figure on the keys to be struck.

To obtain the second and all succeeding root figures, proceed the same as for the third root figure illustrated in the method and example given above.

The following will show how it can be done without finding what the product of the number is or what the square of the number is and thereby saving much time. We have not space in which to illustrate one of each class of such equations but will illustrate examples in two classes which will suffice to give the principle of doing it.

(2742 × 58)—(864×74). After multiplying 2742 by 58, 159036 appears as a product. Leave it on the register and deduct one from the multiplier, in this case 74, which leaves 73. Place the finger on the key in the units column which has a red 3 and strike it and each succeeding key to the left in the row bearing red 3's as many times as the corresponding figures of the multiplicand, 864, indicate. Then place the finger on the key in the tens column which bears a red 7 and proceed to strike the row of red 7 keys in the same manner according to the figures of the multiplicand. Next annex as many ciphers to the multiplicand as there are places in the multiplier and subtract it from the amount on the register. In this case there are two places in the multiplier; therefore, subtract 86400, when the answer, 95100, appears on the register.

The hypothenuse of a triangle is 1278 feet; the base is 473 feet; what is the perpendicular? The equation is $$\sqrt{1278^2-473^2}$$. Squaring 1278 we have 1633284. Leaving it on the register, we multiply 473 by 473 according to the red figures. as illustrated in Example 1, and then annex three ciphers to 473, and subtract it, when 1409555 appears on the register. Extracting the square root, we have 1187, +586. If you desire to express the root as a mixed number, all that is necessary is to multiply the root figures already obtained by 2 and add 1. Then place the result under the remainder as a denominator. In this case it would be as follows: 1187×2=2374. 2374+1=2375; therefore the root is 1187 586.2375.

It is evident that the periods of ciphers can be annexed to the remainder and the root carried out decimally if desired.