To determine the number of figures in the result by rough cancelling and mental calculation, we note that 4·71 enters 432 about 100 times; 9·8 enters 17·5 about 2; 6·21 into 32·4 about 5; and 2·17 into 4·12 about 2. This gives (500)/(4) = 125, showing that the result contains 3 digits. From the slide rule we read 141, which is therefore the result sought. The occasional traversing of the slide through the rule, to interchange the indices—a contingency which the use of the C and D scales always involves—may often be avoided by a very simple expedient. Such an example as (6·19 × 31·2 × 422)/(1120 × 8·86 × 2.09) = 3·93 is sometimes cited as a particularly difficult case. Working through the expression as given, two traversings of the slide are necessary; but by taking the factors in the slightly different order, (6·19 × 31·2 × 422)/(8·86 × 2·09 × 1120), _so that the significant figures of each pair are more nearly alike_, we not only avoid any traversing the slide, but we also reduce the extent to which the slide is moved to effect the several divisions. Such cases as (_a_ × _b_)/(_c_ × _d_ × _e_ × _f_ × _g_) or (_a_ × _b_ × _c_ × _d_ × _e_)/(_f_ × _g_) really resolve themselves into (_a_ × _b_ × 1 × 1 × 1)/(_c_ × _d_ × _e_ × _f_ × _g_) and (_a_ × _b_ × _c_ × _d_ × _e_)/(_f_ × _g_ × 1 × 1 × 1), but, of course, if rules are used to locate the decimal point, the 1’s so (mentally) introduced are not to be counted as additional figures in the factors. MULTIPLICATION AND DIVISION WITH THE SLIDE INVERTED. If the slide be inverted in the rule but with the same face uppermost, so that the Ɔ scale lies adjacent to the A scale, and the right and left indices of the slide and rule are placed in coincidence, we find the product of any number on D by the coincident number on Ɔ (readily referred to each other by the cursor) is always 10. Hence, by reading the numbers on Ɔ as decimals, we have over any unit number on D, its _reciprocal_ on Ɔ. Thus 2 on D is found opposite 0·5 on Ɔ; 3 on D opposite to 0·333; while opposite 8 on Ɔ is 0·125 on D, etc. The reason of this is that the sum of the lengths of the slide and rule corresponding to the factors, is always equal to the length corresponding to the product—in this case, 10.