Speed Mathematics Simplified (28 page)

Two: 7 x 7 ends in 9. 4 x 7 is in the 20's. 9 plus 2 (complement, slash) is 1, from 1 is 0:

Three: 4 x 7 ends in 8, and 8 from 4 is (complement) 6 and slash.

BUT—when we slash a zero, we must always slash the digit to the left of it as well. The 2 to the left of the 0 is already slashed, but we slash it again. We have no choice. If one slash reduces a digit in value by one, two slashes reduce it in value by two, leaving nothing of that double-slashed 2:

Why does it happen this way? The answer is simply that we are multiplying and subtracting at the same time, digit by digit. The use of a complement in either case calls for recording a ten in multiplying (which means canceling a ten here, since we put down only the result of the subtraction) or canceling a ten in subtraction. Now and then, both may affect the same digit—as they did here.

That slashed 0, remember, is now a 9.

Let's go on. See the next answer digit as 8 into 96—MORE THAN TEN.

This signals the need for raising the answer digit by 1. This is not difficult. Just underline it, subtract the divider from the remainder, and the problem now looks like this:

The remainder 222 comes from the subtraction of 74 from 96—really, of course, 740 from 962.

Now we will find the last digit of the answer. 8 into 22 is 3. Multiply and subtract at the same time on your pad and find out whether or not this problem comes out even.

Do it yourself.

Does the problem come out even or not?

Longer Dividers

Perhaps you are already wondering whether dividing by numbers of three or more digits make things much more complicated.

The answer is, not much.

Nothing in our technique changes one bit, except that we repeat step two of no-carry multiplication as many times as we need to in order to get a full subtraction. Dividing by four- or five-digit dividers is no harder than dividing by two-digit dividers. There are more details and it will take a little longer, but the process is not really different.

You still divide by only the first digit of your divider, raised in value by one, to produce automatically each succeeding digit of your answer. Should that answer digit need revising, that fact will be signaled to you when the remainder is larger than the divider. If you miss that signal, you are notified again when the next answer digit seems to be ten or more.

Once in a very great while an answer digit will need revising twice. After you have raised it once and adjusted your remainder, the remainder will
still
be larger than the divider—and the next answer digit will
still
be ten or more. In such rare cases, just underline the answer digit again (raising it in value by 2) and subtract the divider once again from the remainder before continuing.

The following example is admittedly a wild extreme, so obvious on the face of it that even by rote you could hardly get into this sort of situation. Yet, even should you abandon all your number sense and follow every rule without looking at the problem itself, the rules would eventually bail you out. You would have to raise your answer digit no less than five times, yet it would ultimately be right:

This, as we said, is absurd. Yet it demonstrates the absolute reliability of the operating rules even when your own common sense sees nothing wrong. You divide 2 into 9 and see an answer digit of 4, so your first remainder is 55. You raise the answer by one and subtract the divider, giving a remainder of 44. This goes on through four more revisions of the answer digit, the last one taking care of the remainder 11.

The problem was selected especially to demonstrate this ultimate possibility. If an answer digit seems to need revising more than once or twice, sit back and look at the problem as a whole. Chances are you have overlooked something very obvious. The rules are as important to fast mathematics as trees are to a forest—but we take note of the forest first, then use the trees. The folk saying is too obvious to need repetition here.

One more point might be mentioned. If you work entirely by rote, you might sometimes be confused by the placement of your remainder when working out a problem like this:

The first answer digit, by inspection, is 2—4 into 8. But beware when you start to develop the remainder:

One: 3 x 2 is in the 0's, and 0 from—

STOP. 0 from 8? This cannot be so. The answer digit would need several revisions. No, in this case it is 0 from 0—the unshown 0 to the left of 8.

Why? Your own number sense should give you a strong inkling. To distill it into an operating rule, it is because you divided into the first
one
digit of the number divided, rather than into the first
two
digits. This in effect moves the product of this first multiplication one place to the left.

So when you divide the first digit of the divider into the first digit only of the number divided, start right out by ignoring what that first answer digit and the first digit of the divider would “be in.” They wouldn't “be in” anything but the zeros, or you would have divided into the first
two
digits of the number divided.

The rest of this particular problem, by the way, shows a typical example of two-digit revisions. Let's go through it. So far, the figures look like this:

The next answer digit, by inspection, is 6—4 into 25. Put it down and work out the remainder:

One: 3 x 6 is in the 10's, and 1 from 2 is 1.

Two: 3 x 6 ends in 8. 2 x 6 is in the 10's. 8 and 1 is 9, and 9 from 5 is—complement, slash:

Three: 2 x 6 ends in 2, and 2 from 6 is 4:

Look at the remainder. Your growing number sense might show right away that it is exactly twice the divider. If not, you would at least notice that it is larger than the divider, so you underline the 6 to raise it to 7, and subtract the divider from the remainder: 3 from 6 is 3. 2 from 4 is 2.

Look at the remainder again. It is the same as the divider. Underline the underlined 6 once again, subtract the divider, and the problem comes out even. The answer is 2 6, which you read or rewrite as 28.

Get out your pad now, have an absolutely clean page on top, and start one problem with a five-digit divider. This one problem embodies just about every possible wrinkle in shortcut division.