Part 5

EX.—79 × 91 = 7189.

In this case the division line 91 on C indicates on D that the answer lies between 7180 and 7190. As the last figure must be 9, it is at once inferred that the last two figures are 89.

When there are more than three figures in either or both of the factors, the fourth and following figures to the right must be neglected. It is well to note, however, that if the first neglected figure is 5, or greater than 5, it will generally be advisable to increase by 1 the third figure of the factor employed. Generally it will suffice to make this increase in one of the two factors only, but it is obvious that in some cases greater accuracy will be obtained by increasing both factors in this way.

CONTINUED MULTIPLICATION.—To find the product of more than two factors, we make use of the cursor to mark the position of successive products (the value of which does not concern us) as the several factors are taken into the calculation. Setting the index of C to the 1st factor on D, we bring the line of the cursor to the 2nd factor on C, then the index of C to the cursor, the cursor to the 3rd factor, index of C to cursor, and so on, reading the final product on D under the last factor on C. (Note that the 1st factor and the result are read on D; all intermediate readings are taken on C.)

If the rule for the number of digits in a product is used, it is necessary to note the number of times multiplication is effected with the slide projecting to the right. This number, deducted from the sum of the digits of the several factors, gives the number of digits in the product. Ingenious devices have been adopted to record the number of times the slide projects to the right, but some of these are very inconvenient. The author’s method is to record each time the slide so projects, by a minus mark, thus −. These can be noted down in any convenient manner, and the sum of the marks so obtained deducted from the sum of the digits in the several factors, gives the number of digits in the product as before explained.

EX.—42 × 71 × 1·5 × 0·32 × 121 = 173,200.

The product given, which is that read on the rule, is obtained as follows:—Set R.H. index of C to 42 on D, and bring the cursor to 71 on C. Next bring the L.H. index of C to the cursor, and the latter to 1·5 on C. This multiplication is effected with the slide to the right, and a memorandum of this fact is kept by making a mark −. Bring the R.H. index of C to the cursor and the latter to 0·32 on C. Then set the L.H. index of C to the cursor and read the result, 1732, on D under 121 on C, while as a slide again projects to the right, a second − memo-mark is recorded. There are 2 + 2 + 1 + 0 + 3 = 8 digits in the factors, and as there were 2 − marks recorded during the operation, there will be 8 − 2 = 6 digits in the product, which will therefore read 173,200 (173,194·56).

For a very rough evaluation of the result, we note that 1·5 × 0·3 is about 0·5; hence, as a clue to the number of figures we have

40 × 70 × 60 = 3000 × 60 = 180,000.

DIVISION.

The instructions for multiplication having been given in some detail, a full discussion of the inverse process of division will be unnecessary.

RULE FOR DIVISION.—Place the divisor on C, opposite the dividend on D, and read the quotient on D under the index of C.

EX.—225 ÷ 18 = 12·5.

Bringing 18 on C to 225 on D, we find 12·5 under the L.H. index of C.

As in multiplication, the factors are treated as whole numbers, and the position of the decimal point afterwards decided according to the following rule, which, as will be seen, is the reverse of that for multiplication:—

RULE FOR THE NUMBER OF DIGITS IN A QUOTIENT.—If the quotient is read with the slide projecting to the LEFT, subtract the number of digits in the divisor from those in the dividend; but if read with the slide to the RIGHT, ADD 1 to this difference.

In the above example the quotient is read off with the slide to the right, so the number of digits in the answer = 3 − 2 + 1 = 2.

EX.—0·000221 ÷ 0·017 = 0·013.

Here the number of digits in the dividend is −3, and in the divisor −1. The difference is −2; but as the result is obtained with the slide to the right, this result must be increased by 1, so that the number of digits in the quotient is −2 + 1 = −1, giving the answer as 0·013.

If preferred, the result can be obtained in the manner referred to when considering the multiplication of decimals. Thus, treating the above as whole numbers, we find that the result of dividing 221 by 17 = 13, since the difference in the number of digits in the factors, which is 1, is, owing to the position of the slide, increased by 1, giving 2 as the number of digits in the answer. Then by the rules for the division of decimals we know that the number of decimal places in the quotient is equal to 6 − 3 = 3, showing that a cypher is to be prefixed to the result read on the rule.

As in multiplication, so in division, we have a

GENERAL RULE FOR NUMBER OF DIGITS IN A QUOTIENT.—When the first significant figure in the DIVISOR is greater than that in the DIVIDEND, the number of digits in the quotient is found by subtracting the digits in the divisor from those in the dividend. When the contrary is the case, 1 IS TO BE ADDED to this difference. When the first figures are the same, those following must be compared.

ESTIMATION OF THE FIGURES IN A QUOTIENT.—The method of roughly estimating the number of figures in a quotient needs little explanation.

EX.—3·95 ÷ 5340 = 0·00074.

Setting 534 on C to 3·95 on D we read under the (R.H.) index of C, the significant figures on D, which are 74. Then 3·9 ÷ 5 is about 0·8 and 0·8 ÷ 1000 gives 0·0008 as a rough estimate.

EX.—0·00000285 ÷ 0·000197 = 0·01446.

Regarding this as 2·85 × 10^{−6} ÷ 1·97 × 10^{−4} we divide 2·85 by 1·97 and obtain 1·446. Dividing the powers of 10 we have 10^{−6} ÷ 10^{−4} = 10^{−2}, so the decimal point is to be moved two places to the left and the answer is read as 0·01446.

Another method of dividing deserves mention as of special service when dividing a number of quantities by a constant divisor:—Set the index of C to the divisor on D and over any dividend on D, read the quotient on C.