_Number of Digits in Result in Combined Multiplication and Division._—For those who use rules the author’s method of determining the decimal point in combined multiplication and division may be used. Each time _multiplication_ is performed with the slide projecting to the _right_, make a − mark; each time _division_ is effected with the slide to the right, make a | mark; _but allow the_ | _marks to cancel the_ − _marks as far as they will_. Subtract the sum of the digits in the denominator from the sum of digits in the numerator, and to this difference _add_ any uncancelled memo-marks, if of | character, or _subtract_ them if of − character. EX.—(43·5 × 29·4 × 51 × 32)/(27 × 3·83 × 10·5 × 1·31) = 1468. [Sidenote: ⵜ ⵜ ⵏ ⵏ] Set 27 on C to 43·5 on D, and as with this _division_ the slide is to the right, make the first ⵏ mark. Bring cursor to 29·4 on C, and as in this _multiplication_ the slide is to the right, make the first − mark, cancelling as shown. Setting 3·83 on C to the cursor, requires the second ⵏ mark, which, however, is cancelled in turn by the multiplication by 51. The division by 10·5 requires the third ⵏ mark, and after multiplying by 32 (requiring no mark) the final division by 1·31 requires the fourth ⵏ mark. Then, as there are 8 numerator digits, 6 denominator, and 2 uncancelled memo-marks (which, being 1, are additive) we have Number of digits in result = 8 − 6 + 2 = 4. Had the uncancelled marks been − in character, the number of digits would have been 8 − 6 − 2 = 0. For quantities less than 0·1 the digit place numbers will be _negative_. The troublesome addition of these may be avoided by transferring them to the opposite side and treating them as positive. _2_ _4_ 0·00356 × 27·1 × 0·08375 Thus:— ───────────────────────── = 288 0·1426 × 9·85 × 0·00002 _2_ _1_ _1_ The first numerator, 0·00356, has −2 digits. Note this by placing 2 _below the lower line_ as shown. 27·1 has 2 digits; place 2 over it. 0·08375 has −1 digit; hence place 1 _below the lower line_. The first denominator has no digits; the second, 9·85, has 1 digit; hence place 1 under it. 0·00002 has −4 digits; place 4 _above the upper line_. The sum of the top series is 2 + 4 = 6; of the bottom series 2 + 1 + 1 = 4. Subtracting the bottom from the top, we have 6 − 4 = 2 digits, to which 1 has to be added for an uncancelled memo-mark, and the result is read as 288. Moving the decimal point often facilitates matters. Thus, (32·4 × 0·98 × 432 × 0·0217)/(4·71 × 0·175 × 0·00000621 × 412000) is much more conveniently dealt with when re-arranged as (32·4 × 9·8 × 432 × 2·17)/(4·71 × 17·5 × 6·21 × 4·12) = 141.