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Authors: Edward Stoddard

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This problem is bound to be a bit tedious, in any method of arithmetic. We can get a rapid estimate, as the next chapter will show, but for a complete answer there is no avoiding a number of steps.

Take a deep breath and plunge in.

4 into 18—our way of producing the first answer digit automatically—is 4. This 4 goes over the sixth digit of the number divided, since there are five digits in the divider and we started by dividing into two digits rather than one. Now we start on the remainder:

One: 3 x 4 is in the 10's, and 1 from 1 is 0.

Two: 3 x 4 ends in 2. 6 x 4 is in the 20's. 2 plus 2 is 4, and 4 from 8 is 4.

(We are still spelling out every step to make it clear. But as you practice the smooth handling of these steps, your eye and mind should begin to jump from one fact to the other almost without intermediate thought. Step two above will, with experience, become “2—4—4.”)

Three: 6 x 4 ends in 4. 1 x 4 is in the O's. 4 from 9 is 5.

Four: 1 x 4 ends in 4. 8 x 4 is in the 30's. 4 and 3 are 7, and 7 from 2 is (complement) 5 and slash left.

Five: 8 x 4 ends in 2. 2 x 4 is in the O's. 2 from 6 is 4.

Six: 2 x 4 ends in 8. 8 from 8 is 0.

If you followed these steps on your pad, here is what your work should look like:

Are you ready to go on? Or would it be a good idea to take another look at the remainder? The remainder is larger than the divider. So underline the 4—raising it to 5—and subtract 36182 from 44640, left to right, canceling in the answer:

If
is smaller than 36182, we are ready to go on. Divide 4 into 8 and put down 2 as the second answer digit. Start developing the remainder:

One: 3 x 2 is in the 0's. Ignore it, because we divided into one digit rather than two.

Two: 3 x 2 ends in 6. 6 x 2 is in the 10's. 6 plus 1 is 7, and 7 from 8 is 1.

Three: 6 x 2 ends in 2. 1 x 2 is in the 0's. 2 from 3 is 1.

Four: 1 x 2 ends in 2. 8 x 2 is in the 10's. 2 plus 1 is 3, and 3 from 5 is 2.

Five: 8 x 2 ends in 6. 2 x 2 is in the 0's. 6 from 8 is 2.

Six: 2 x 2 ends in 4. 4 from 2 (complement) 8, and slash.

Check your jottings against the model at this point:

This should be enough step-by-step explanation. Go ahead and finish this problem. You will find that it does not come out even. There will be a remainder. Make sure to examine the remainder carefully. There is a reason why you should.

Do not look ahead to the finished problem until you have a remainder that satisfies you. Then, if you feel you understand all the techniques in this chapter, go ahead to build speed in division.

The remainder, as you undoubtedly discovered, held within it a revision of the last digit:

Final answer, 523; remainder, 3,638.

9

BUILDING SPEED IN DIVISION

T
HE last exercise in the last chapter was a stunner. It was, just from the quantity of digits to be handled, the most tedious situation you are likely to face in arithmetic. The numbers go on and on, and if you need a full answer there is simply no way to avoid dealing with every single digit.

Most division is much simpler.

First of all, we seldom need to work out any division problem of this length in such detail. Largely because American business has become accustomed (and wisely so) to dealing in rounded-off numbers, you would most likely find such a problem rounded off to start with. Second, the only reason we had to use every single digit was to get a fully accurate remainder. We would have got precisely the same whole-number answer by cutting down the divider from five to three digits.

Remember the two reasons for working from left to right: it is more natural, and it is also self-estimating. Turn back to the last problem for a moment, then compare the final working with this version of
the same problem:

This is really the same problem—but a rounded-off version of it. We rounded it off before beginning by doing two things:

First, we rounded off the divider to three digits because we saw simply by inspection that the whole-number answer would be in three digits. In rounding off, the first three digits (361) became 362 because the following digit is 5 or more.

Second, we dropped the
same
number of digits from the number divided as we did from the divider. This ensures that our answer will not be ten times too big or ten times too small.

Rounding off such a problem is obviously faster as well as simpler. For most purposes, it is quite accurate enough. You will note that the remainder is not the same, and sometimes the last digit itself might be off by one or two points in value—but we are still more accurate than a slide rule.

When you need a very quick estimate, you can carry this even further. If you care only about the first two digits of the answer, then round off your divider to two digits and cross out as many digits in the number divided as you did in the divider.

In this event, the full divider 36182 becomes 36. The number divided, 18936824, becomes the far more manageable 18937.

Your solution now looks like this:

Notice that the third digit of your answer is no longer accurate at all. But your first two digits are.

You would seldom simplify a problem to quite this extent, because the possible error is ten to twenty per cent, but it is a useful device to know when speed rather than perfect accuracy is required for a very fast approximation.

Try rounding off one sample to make sure you have the idea firmly in mind—especially the proper handling of the number divided. Reduce this example to a form suitable for a three-digit answer:

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