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Overview
With no math background required and no long list of rules to memorize, Doing Simple Math in Your Head teaches you how to simplify math problems, provides ample reallife practice problems and solutions, and gives you the necessary background in basic arithmetic to handle everyday problems quickly. Math is finally made easy with the right shortcuts, tips, and strategies.
Product Details
ISBN13:  9780962734151 

Publisher:  Coast Publishing 
Publication date:  02/01/1992 
Pages:  144 
Product dimensions:  8.52(w) x 5.55(h) x 0.40(d) 
About the Author
W. J. Howard is a mathematician who specializes in making technical information understandable to the general reader.
Read an Excerpt
Doing Simple Math in Your Head
By W. J. Howard
Chicago Review Press Incorporated
Copyright © 1992 W. J. HowardAll rights reserved.
ISBN: 9781556524233
CHAPTER 1
MAKING THINGS EASIER
You can simplify problems in various ways: ...
Numbers are not rigid, static things; they're malleable and fluid, and can be shaped to meet your needs. A problem comes up involving the number 15. You don't have to work with it as such. If it suits you, you can think of it as 10 + 5, or 5 × 3, or 30/2 — whatever fits your need. How do you know what will fit? You don't, without a little mental experimenting. And in experimenting a guiding light is the magic number 10, as you'll see.
A wide variety of problems will succumb to just a few easytouse methods: reordering numbers, rearranging them, breaking them up, using equivalents and identities, and approximating and rounding off. Always, the focus is on the number 10; it and its multiples are the easiest numbers to handle. As you review these methods, don't try to memorize; understand the logic ... and then practice. When ready, turn to Chapter 2 for practice, or to Chapter 3 for help in understanding.
by reordering numbers; ...
Numbers come at you helter skelter. You want to add 14, 39 and 6. (Perhaps you're in a store and thinking of buying some items. In this chapter we won't worry about why you want to add, multiply, etc. — there will be many examples in the problems later.) The first thing to do is to reorder them to (14 + 6) + 39. Now 14 + 6 gives you 20 — a multiple of 10 — and 20 is easily added to 39. Result, 59 — which you can do in your head in much less time than it takes to tell about it.
You have three numbers to multiply: 25 × 33 × 8. Don't try to multiply 25 × 33 (this in itself is not too difficult, but you'd still have to multiply by 8); first multiply 8 × 25, getting 200 (aha!, a multiple of 10), and then multiply this by 33. Answer: 6,600.
Maybe a problem involves a fraction: (¾) × 48. You need to multiply 48 by 3 and divide by 4. But by all means, divide first!
by adding and subtracting a number; ...
You want to add 87 and 36. Notice that 87 is close to 90 — a multiple of 10. Change 87 to 90 by adding 3, and then, to compensate, subtract that same 3 from 36, getting 33. Rearranging a problem by adding and subtracting the same number doesn't change the answer. So the original problem of 87 + 36 is now 90 + 33, which is easier. You can make it even easier by adding and subtracting 10: adding 10 to 90 gives 100 — a beautiful multiple of 10 — and subtracting 10 from 33 gives 23. So the original problem has now been changed to 100 + 23. It doesn't look anything like the original problem — in fact it's now no problem at all — but it will give the same answer.
In subtraction, try to change the number being subtracted to a multiple of 10. To subtract 19 from 44, add 1 to 19, giving 20, and add that same number (1) to 44, giving 45. The rearranged problem is now easy: 45  20, or 25.
by multiplying and dividing by a number; ...
Multiplying or dividing by 2 is easy to do in your head — a fact that has many benefits. To multiply by 5, think of multiplying by 10 and taking half of that. This gives the same result and is often easier. Suppose the problem is 5 × 14.6. Multiplying and dividing by 2 changes it to (2 × 5) × (14.6/2), so we're now multiplying by 10 (that is, 2x5) and dividing by 2. The result is 146/2, or 73. This also works for numbers ending in 5; multiply by 2 to get a multiple of 10, and divide by 2 to compensate. For example, 15 × 26 is changed to (2 × 15) × (26/2), or 30 × 13, or 390. We divided by 2 before multiplying here, whereas above we did so after — do whatever makes the problem easier.
When you see 2.5 or 25, think of multiplying and dividing by 4, so you'll be multiplying by 10 or 100 (and dividing by 4). A forbiddinglooking 2.5 × 68 is changed to (4 × 2.5) × 68/4, or 10 × 17, or 170. (Also see fraction/decimal equivalents, later.)
by breaking up numbers ...
Many problems give you a choice of how to proceed. Don't waste time waffling, or you'll defeat the purpose of making things simple. Consider 12 × 19. Looking for a 10 multiple, you think of 12 as 10 + 2, so the problem is now (10 + 2) × 19, which is (10 × 19) + (2 × 19), or 190 + 38, or 228. (If you have a question about this, see p. 95.) Alternatively, for 12 × 19 you might have thought of 19 as 20  1, since 20 is a multiple of 10. The problem then becomes 12 × (20  1), or (12 × 20)  (12 × 1), or 240  12, or 228 again. If you have trouble with 240  12, remember to change the problem so you're subtracting a multiple of 10. In this case, subtract 2 from each number: 240  12 = (240  2)  (12  2), or 238  10, which is 228.
On the previous page we multiplied 15 × 26 by multiplying by 2 and dividing by 2. Instead, we might have broken up 15 into 10 + 5; the problem becomes (10 + 5) × 26, which is 260 + 130, or 390.
in various ways, ...
The numbers 9 and 11 are easy to handle, since they're both so close to 10; think of 9 as 10  1 and 11 as 10 + 1. To multiply by 9, first multiply the number by 10 and then subtract the number: 9 × 26 = (10  1) × 26 = (10 × 26) (1 × 26), which is 260  26. Some may wish to simplify even further by adding 4 to each number: 260  26 = (260 + 4)  (26 + 4), which is 264  30, or 234.
To multiply by 11, multiply by 10 and add the number: 11 × 26 = (10 + 1) × 26, which is (10 × 26) + (1 × 26), or 260 + 26, or 286.
You can even handle larger numbers, like 99 and 101, since they're so close to 100. For example, 99 × 84 = (100  1) × 84, or (100 × 84)  (1 × 84), or 8,400 84. You can simplify to (8,400 + 16)  (84 + 16), or 8,416  100, or 8,316. If the problem is 101 × 84 it's even easier: 101 × 84 = (100 + 1) × 84, or (100 × 84) + (1 × 84), or 8,400 + 84, or 8,484.
including factoring; ...
When multiplying or dividing, it's often useful to factor numbers (composite numbers can be factored into their components: 6 = 2 × 3, for example — see p. 97). Suppose you want to divide 78 by 6. You can perhaps do this directly in your head, but if you have difficulty, divide first by 2 and then by 3. Dividing by 2 gives you 39, and dividing this by 3 gives you the answer, 13.
Another problem: divide 396 by 18. Since 18 = 2 × 3 × 3, you can divide first by 3, getting 132, then by 2, getting 66, then by 3 again, getting the final answer, 22. You might also solve this directly by noting 20 × 18 is 360, and there's still 2 × 18 to go.
Suppose you want to multiply 28 × 13. Since 28 = 7x2x2, you can restate this problem as (7 × 2 × 2) × 13, which is 7 × (2 × 26), or 7 × 52. For 7 × 52 you can break up 52 into 50 + 2 and multiply separately: (7 × 50) + (7x2) is 350 + 14, or 364.
by using fraction/decimal equivalents ...
Fractions are surprisingly useful, particularly in multiplication problems (see p. 114). To multiply 50 × 38, it's easy to multiply 100 × 38 and take half of that — 1,900. Another problem: 0.25 × 436. You'd have quite a time multiplying this in your head. But realizing that 0.25 is ¼, you just divide 436 by 4 and have the answer, 109.
Once you become familiar with fractions and their decimal equivalents, you find many opportunities to use them. It's not just in the form of numbers less than one that fractions are useful. If the second problem above had been 2.5 × 436, you'd do the same thing — divide by 4 — but then, since 2.5 is 10 times 0.25, you'd have to multiply by 10 to get the answer: 1,090. With practice, you get in the habit of looking for numbers that correspond to fractional equivalents, knowing you can multiply or divide by a power of 10 (10,100, etc. — see p. 98) to make the answer come out right.
to convert awkward numbers ...
Some fractional equivalents are familiar; it's common knowledge that ½ = 0.5, ¼ = 0.25, and ¾ = 0.75. Also, 1/3 = 0.33 and 2/3 = 0.67 (to two places). A problem involving these numbers can often be handled quickly. For example: 67 × 39 seems at first like a difficult problem. But think of 67 as (2/3) × 100. Note also that 1/3 of 39 is 13, so 2/3 is 26, and you have an approximate answer right away: (2/3) × 100 × 39 is 2,600. The answer is not exact since (2/3) × 100 is not exactly 67.
As for the less familiar equivalents: since you know that ¼ is 0.25, you also know that 1/8, being half of ¼, is 0.125. So if a problem involves 125, you'll think of dividing by 8 and multiplying by the proper power of 10. By dividing by 2 again, you also know that 1/16 is 0.0625. Or take 1/3, which you know is 0.333; 1/6 is half of that, or 0.167, and 1/12 is half of that, or 0.083. Any or all of these relations can come in handy ... if you think of them.
to simpler ones; ...
You can do some difficultlooking problems in your head using fraction/decimal equivalents. How about 12.5 × 17? Since 1/8 is 0.125,12.5 is (1/8) × 100, and 12.5 × 17 is (1/8) × 100 × 17. Now 17/8 is 2 1/8, or 2.125. So 12.5 × 17 is 2.125 × 100, or 212.5. It takes longer to describe it than to do it. Notice that this works even though 8 does not evenly divide 17. Another: 56 ÷ 17.5 looks impossible until you realize 17.5 = 10 × 1.75, and 1.75 = 7/4. So (see p. 106) 56 ÷ 17.5 = 56 × (4/7) × (1/10), which is (8 × 4)/10, or 3.2. Not an easy problem.
If you multiply 200 × 300, you'll first think "2 × 3 = 6," then start wondering how many zeros to tack on. This is how to handle a problem using fraction/decimal equivalents. Think about the numbers or digits first, without worrying about size of the answer. This lets you make useful associations. Size is just a matter of factors of 10 — how many zeros or where the decimal point goes.
and by using identities ...
Try multiplying 5  3 by 5 + 3. Yes, the answer will be 2 × 8, or 16, but let's multiply it out: (5  3) × (5 + 3) = (5 × 5) + (5 × 3)  (5 × 3)  (3 × 3). Since the middle terms cancel, this is just (5 × 5)  (3 × 3), or 25  9, or 16. To save space, it's usual to write 5 × 5 and 3 × 3 as 52 and 32, respectively. So we've found that (5  3) × (5 + 3) = 52  32.
We don't need this kind of help in multiplying 2 by 8, but no matter what the two numbers are, the same relation will hold. For any two numbers, say a and b, (a  b) × (a + b) = a2  b2 — as you can see by multiplying out. This relation is very useful for larger numbers. (We've sneaked in a little algebra here — using letters for numbers.)
Say you want to multiply 17 × 23. Looking for a multiple of 10, you see they're both near 20, so you can think of this as (20  3) × (20 + 3). And this is simply 202  32, or 400  9, or 391.
that apply to any numbers.
Flushed with success at finding one useful relation, let's look for another. Rather than (5  3) × (5 + 3), suppose we had (5 + 3) × (5 + 3); what then? Multiplying out, we have (5 × 5) + (5 × 3) + (5 × 3) + (3 × 3). This time, the middle terms don't cancel, so we're left with 52 + 2 × (5 × 3) + 32, or 25 + 30 + 9, or 64 — what we expected, since 82 = 64. We've found that (5 + 3) 2 = 52 + 2 × (5 × 3) + 32.
Again, 5 and 3 are not unique — this same relation holds for any two numbers a and b: (a + b) 2 = a2 + 2 × (a × b) + b2 — no big deal, but still useful. For example, you want to "square" 41 (that is, multiply 41 × 41). Think of it as (40 + 1) 2, which is 402 + 2 × 40 × 1 + 12, or 1,600 + 80 + 1, or 1,681.
What about (a  b)2? Multiplying out, you see that (a  b)2 = a2  2 × (a × b) + b2. So, for example, 282 = (30  2)2 = 302  2 × (30 × 2) + 22, or 900  120 + 4, or 784. Another tool to add to our repertoire.
Round off and approximate ...
With each digit worth ten times the digit to its right, our decimal system makes it easy to focus on the ones that count. But you don't want to forget entirely about the righthand digits; if you go through life thinking of $1.75 as $1, you're in trouble. So what you do is "round off" (see p. 113). To round off a number, drop unwanted digits on the right. If the first digit dropped is 5 or greater, increase the preceding digit by 1; otherwise leave it unchanged. For example, to three places, 1/3 is rounded to 0.333, and to two places, 0.33; 2/3 is rounded to 0.667 and 0.67.
Other examples: $1.75, rounded to the nearest dime, is $1.80, since the digit dropped is 5; to the nearest dollar, $1.75 rounds to $2. A bill of $12.87 (next page) rounds to $12.90 to the nearest dime, $13 to the nearest dollar, and $10 to the nearest ten dollars. Rounding off just makes clear what we mean when we say $12.87 is "about $13."
where appropriate, ...
If you're figuring how much to tip in a restaurant, you certainly don't need to work it out to the last penny — unless you want to establish a reputation as a pennypincher, even while trying to be generous with a tip. Just an approximate amount is all that's expected. Often you want to give a 15% tip. The easiest way to do this is to first figure 10% by moving the decimal point one place to the left (see pp. 110 and 116), then take half of this, which will be 5%, and add the two. Round off as you proceed. Suppose your bill is $12.87; 10% of this is about $1.30, and half of $1.30 is $0.65. Adding the two, you have $1.95. So if you leave about $2 you'll be doing what you wanted to do.
Many items are priced at just less than an even dollar figure, for sales purposes: $8.99, $13.95, and so on. You don't usually care about the pennies, so just round these up to the nearest dollar: $8.99 + $13.95 is about $9 + $14, or $23.
but be aware that errors can creep in.
It's easier, of course, to work with numbers when you approximate and round off — that's why you do it. But you pay a penalty, as your answers are less accurate. So do it when accuracy is not at a premium. For example, in adding 873 and 465, you might use 900 and 500, getting 1,400. The correct answer of 1,338 is quite a bit different. Does it matter? You must be the judge.
Larger errors can be made in multiplying. For 67 × 31, you might try 70 × 30, or 2,100. Not bad: the correct answer is 2,077. But suppose you simplified 72 × 26 to 70 × 30. The result, 2,100 again, is quite a bit larger than the correct answer, 1,872. There is too much change in the smaller number: 26 became 30. Instead of adding 26 72's together (p. 84), you're adding 30, or four more of the larger number than there should be. The principle: be careful in rounding off numbers before multiplying — particularly the smaller of two numbers.
Ready for problems? Mental exercises help.
Doing math in your head is unfamiliar ground; even with the suggestions given, you may have trouble. One thing you can do to feel more comfortable is to use idle moments to practice. First, a logical extension to the "tables" you learned long ago is to add and multiply twodigit numbers. Start with easier ones like 11,12 and 13 and work up. Try doing it in different ways, using the suggestions in this chapter and perhaps some of your own. Next, become familiar with fraction/decimal equivalents, for fractions with denominators ranging from 2 to 12. Don't try to memorize; look for relations (such as 5/12 = 3/12 + 2/12, which is ¼ + 1/6, or 0.25 + 0.167, or 0.417). Finally, devise problems for yourself, such as converting monthly to annual bills or salaries. You can do 12 × $750 in three ways: (10 + 2) × $750,12 × ($700 + $50), and (¾) × 12 × $1,000. After a while, you'll find it so easy and satisfying you'll wonder why you haven't been using your head more all along.
CHAPTER 2PROBLEMS AND SOLUTIONS
The usual "book" problems don't suit our purpose, ...
Problems can be stated in different ways; shown here are three common types. You'll find many problems of Type 1 in arithmetic books. We've all done these and gained some good practice with numbers. You also find problems of Type 2, where the numbers are not lined up vertically and are a little harder to work with. And then there are "word problems" — Type 3 — which try for realism.
As mentioned in the Introduction, problems that you come across in daily life are not like this. They're certainly different than Types 1 and 2, since you have to decide what to do — there's no +, , × or ÷ sign to tell you. But they also differ from Type 3; real problems often don't even give you numbers, as such, to work with; you have to deal with numbers as words in your mind. Notice that with Type 3, the numbers appear as numbers, not as words. How can problems be stated so they're more like those you'll be doing in your head?
(Continues...)
Excerpted from Doing Simple Math in Your Head by W. J. Howard. Copyright © 1992 W. J. Howard. Excerpted by permission of Chicago Review Press Incorporated.
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by DialABook Inc. solely for the personal use of visitors to this web site.
First Chapter
Doing Simple Math in Your Head
By W. J. Howard
Chicago Review Press Incorporated
Copyright © 1992 W. J. HowardAll rights reserved.
ISBN: 9781556524233
CHAPTER 1
MAKING THINGS EASIER
You can simplify problems in various ways: ...
Numbers are not rigid, static things; they're malleable and fluid, and can be shaped to meet your needs. A problem comes up involving the number 15. You don't have to work with it as such. If it suits you, you can think of it as 10 + 5, or 5 × 3, or 30/2 — whatever fits your need. How do you know what will fit? You don't, without a little mental experimenting. And in experimenting a guiding light is the magic number 10, as you'll see.
A wide variety of problems will succumb to just a few easytouse methods: reordering numbers, rearranging them, breaking them up, using equivalents and identities, and approximating and rounding off. Always, the focus is on the number 10; it and its multiples are the easiest numbers to handle. As you review these methods, don't try to memorize; understand the logic ... and then practice. When ready, turn to Chapter 2 for practice, or to Chapter 3 for help in understanding.
by reordering numbers; ...
Numbers come at you helter skelter. You want to add 14, 39 and 6. (Perhaps you're in a store and thinking of buying some items. In this chapter we won't worry about why you want to add, multiply, etc. — there will be many examples in the problems later.) The first thing to do is to reorder them to (14 + 6) + 39. Now 14 + 6 gives you 20 — a multiple of 10 — and 20 is easily added to 39. Result, 59 — which you can do in your head in much less time than it takes to tell about it.
You have three numbers to multiply: 25 × 33 × 8. Don't try to multiply 25 × 33 (this in itself is not too difficult, but you'd still have to multiply by 8); first multiply 8 × 25, getting 200 (aha!, a multiple of 10), and then multiply this by 33. Answer: 6,600.
Maybe a problem involves a fraction: (¾) × 48. You need to multiply 48 by 3 and divide by 4. But by all means, divide first!
by adding and subtracting a number; ...
You want to add 87 and 36. Notice that 87 is close to 90 — a multiple of 10. Change 87 to 90 by adding 3, and then, to compensate, subtract that same 3 from 36, getting 33. Rearranging a problem by adding and subtracting the same number doesn't change the answer. So the original problem of 87 + 36 is now 90 + 33, which is easier. You can make it even easier by adding and subtracting 10: adding 10 to 90 gives 100 — a beautiful multiple of 10 — and subtracting 10 from 33 gives 23. So the original problem has now been changed to 100 + 23. It doesn't look anything like the original problem — in fact it's now no problem at all — but it will give the same answer.
In subtraction, try to change the number being subtracted to a multiple of 10. To subtract 19 from 44, add 1 to 19, giving 20, and add that same number (1) to 44, giving 45. The rearranged problem is now easy: 45  20, or 25.
by multiplying and dividing by a number; ...
Multiplying or dividing by 2 is easy to do in your head — a fact that has many benefits. To multiply by 5, think of multiplying by 10 and taking half of that. This gives the same result and is often easier. Suppose the problem is 5 × 14.6. Multiplying and dividing by 2 changes it to (2 × 5) × (14.6/2), so we're now multiplying by 10 (that is, 2x5) and dividing by 2. The result is 146/2, or 73. This also works for numbers ending in 5; multiply by 2 to get a multiple of 10, and divide by 2 to compensate. For example, 15 × 26 is changed to (2 × 15) × (26/2), or 30 × 13, or 390. We divided by 2 before multiplying here, whereas above we did so after — do whatever makes the problem easier.
When you see 2.5 or 25, think of multiplying and dividing by 4, so you'll be multiplying by 10 or 100 (and dividing by 4). A forbiddinglooking 2.5 × 68 is changed to (4 × 2.5) × 68/4, or 10 × 17, or 170. (Also see fraction/decimal equivalents, later.)
by breaking up numbers ...
Many problems give you a choice of how to proceed. Don't waste time waffling, or you'll defeat the purpose of making things simple. Consider 12 × 19. Looking for a 10 multiple, you think of 12 as 10 + 2, so the problem is now (10 + 2) × 19, which is (10 × 19) + (2 × 19), or 190 + 38, or 228. (If you have a question about this, see p. 95.) Alternatively, for 12 × 19 you might have thought of 19 as 20  1, since 20 is a multiple of 10. The problem then becomes 12 × (20  1), or (12 × 20)  (12 × 1), or 240  12, or 228 again. If you have trouble with 240  12, remember to change the problem so you're subtracting a multiple of 10. In this case, subtract 2 from each number: 240  12 = (240  2)  (12  2), or 238  10, which is 228.
On the previous page we multiplied 15 × 26 by multiplying by 2 and dividing by 2. Instead, we might have broken up 15 into 10 + 5; the problem becomes (10 + 5) × 26, which is 260 + 130, or 390.
in various ways, ...
The numbers 9 and 11 are easy to handle, since they're both so close to 10; think of 9 as 10  1 and 11 as 10 + 1. To multiply by 9, first multiply the number by 10 and then subtract the number: 9 × 26 = (10  1) × 26 = (10 × 26) (1 × 26), which is 260  26. Some may wish to simplify even further by adding 4 to each number: 260  26 = (260 + 4)  (26 + 4), which is 264  30, or 234.
To multiply by 11, multiply by 10 and add the number: 11 × 26 = (10 + 1) × 26, which is (10 × 26) + (1 × 26), or 260 + 26, or 286.
You can even handle larger numbers, like 99 and 101, since they're so close to 100. For example, 99 × 84 = (100  1) × 84, or (100 × 84)  (1 × 84), or 8,400 84. You can simplify to (8,400 + 16)  (84 + 16), or 8,416  100, or 8,316. If the problem is 101 × 84 it's even easier: 101 × 84 = (100 + 1) × 84, or (100 × 84) + (1 × 84), or 8,400 + 84, or 8,484.
including factoring; ...
When multiplying or dividing, it's often useful to factor numbers (composite numbers can be factored into their components: 6 = 2 × 3, for example — see p. 97). Suppose you want to divide 78 by 6. You can perhaps do this directly in your head, but if you have difficulty, divide first by 2 and then by 3. Dividing by 2 gives you 39, and dividing this by 3 gives you the answer, 13.
Another problem: divide 396 by 18. Since 18 = 2 × 3 × 3, you can divide first by 3, getting 132, then by 2, getting 66, then by 3 again, getting the final answer, 22. You might also solve this directly by noting 20 × 18 is 360, and there's still 2 × 18 to go.
Suppose you want to multiply 28 × 13. Since 28 = 7x2x2, you can restate this problem as (7 × 2 × 2) × 13, which is 7 × (2 × 26), or 7 × 52. For 7 × 52 you can break up 52 into 50 + 2 and multiply separately: (7 × 50) + (7x2) is 350 + 14, or 364.
by using fraction/decimal equivalents ...
Fractions are surprisingly useful, particularly in multiplication problems (see p. 114). To multiply 50 × 38, it's easy to multiply 100 × 38 and take half of that — 1,900. Another problem: 0.25 × 436. You'd have quite a time multiplying this in your head. But realizing that 0.25 is ¼, you just divide 436 by 4 and have the answer, 109.
Once you become familiar with fractions and their decimal equivalents, you find many opportunities to use them. It's not just in the form of numbers less than one that fractions are useful. If the second problem above had been 2.5 × 436, you'd do the same thing — divide by 4 — but then, since 2.5 is 10 times 0.25, you'd have to multiply by 10 to get the answer: 1,090. With practice, you get in the habit of looking for numbers that correspond to fractional equivalents, knowing you can multiply or divide by a power of 10 (10,100, etc. — see p. 98) to make the answer come out right.
to convert awkward numbers ...
Some fractional equivalents are familiar; it's common knowledge that ½ = 0.5, ¼ = 0.25, and ¾ = 0.75. Also, 1/3 = 0.33 and 2/3 = 0.67 (to two places). A problem involving these numbers can often be handled quickly. For example: 67 × 39 seems at first like a difficult problem. But think of 67 as (2/3) × 100. Note also that 1/3 of 39 is 13, so 2/3 is 26, and you have an approximate answer right away: (2/3) × 100 × 39 is 2,600. The answer is not exact since (2/3) × 100 is not exactly 67.
As for the less familiar equivalents: since you know that ¼ is 0.25, you also know that 1/8, being half of ¼, is 0.125. So if a problem involves 125, you'll think of dividing by 8 and multiplying by the proper power of 10. By dividing by 2 again, you also know that 1/16 is 0.0625. Or take 1/3, which you know is 0.333; 1/6 is half of that, or 0.167, and 1/12 is half of that, or 0.083. Any or all of these relations can come in handy ... if you think of them.
to simpler ones; ...
You can do some difficultlooking problems in your head using fraction/decimal equivalents. How about 12.5 × 17? Since 1/8 is 0.125,12.5 is (1/8) × 100, and 12.5 × 17 is (1/8) × 100 × 17. Now 17/8 is 2 1/8, or 2.125. So 12.5 × 17 is 2.125 × 100, or 212.5. It takes longer to describe it than to do it. Notice that this works even though 8 does not evenly divide 17. Another: 56 ÷ 17.5 looks impossible until you realize 17.5 = 10 × 1.75, and 1.75 = 7/4. So (see p. 106) 56 ÷ 17.5 = 56 × (4/7) × (1/10), which is (8 × 4)/10, or 3.2. Not an easy problem.
If you multiply 200 × 300, you'll first think "2 × 3 = 6," then start wondering how many zeros to tack on. This is how to handle a problem using fraction/decimal equivalents. Think about the numbers or digits first, without worrying about size of the answer. This lets you make useful associations. Size is just a matter of factors of 10 — how many zeros or where the decimal point goes.
and by using identities ...
Try multiplying 5  3 by 5 + 3. Yes, the answer will be 2 × 8, or 16, but let's multiply it out: (5  3) × (5 + 3) = (5 × 5) + (5 × 3)  (5 × 3)  (3 × 3). Since the middle terms cancel, this is just (5 × 5)  (3 × 3), or 25  9, or 16. To save space, it's usual to write 5 × 5 and 3 × 3 as 52 and 32, respectively. So we've found that (5  3) × (5 + 3) = 52  32.
We don't need this kind of help in multiplying 2 by 8, but no matter what the two numbers are, the same relation will hold. For any two numbers, say a and b, (a  b) × (a + b) = a2  b2 — as you can see by multiplying out. This relation is very useful for larger numbers. (We've sneaked in a little algebra here — using letters for numbers.)
Say you want to multiply 17 × 23. Looking for a multiple of 10, you see they're both near 20, so you can think of this as (20  3) × (20 + 3). And this is simply 202  32, or 400  9, or 391.
that apply to any numbers.
Flushed with success at finding one useful relation, let's look for another. Rather than (5  3) × (5 + 3), suppose we had (5 + 3) × (5 + 3); what then? Multiplying out, we have (5 × 5) + (5 × 3) + (5 × 3) + (3 × 3). This time, the middle terms don't cancel, so we're left with 52 + 2 × (5 × 3) + 32, or 25 + 30 + 9, or 64 — what we expected, since 82 = 64. We've found that (5 + 3) 2 = 52 + 2 × (5 × 3) + 32.
Again, 5 and 3 are not unique — this same relation holds for any two numbers a and b: (a + b) 2 = a2 + 2 × (a × b) + b2 — no big deal, but still useful. For example, you want to "square" 41 (that is, multiply 41 × 41). Think of it as (40 + 1) 2, which is 402 + 2 × 40 × 1 + 12, or 1,600 + 80 + 1, or 1,681.
What about (a  b)2? Multiplying out, you see that (a  b)2 = a2  2 × (a × b) + b2. So, for example, 282 = (30  2)2 = 302  2 × (30 × 2) + 22, or 900  120 + 4, or 784. Another tool to add to our repertoire.
Round off and approximate ...
With each digit worth ten times the digit to its right, our decimal system makes it easy to focus on the ones that count. But you don't want to forget entirely about the righthand digits; if you go through life thinking of $1.75 as $1, you're in trouble. So what you do is "round off" (see p. 113). To round off a number, drop unwanted digits on the right. If the first digit dropped is 5 or greater, increase the preceding digit by 1; otherwise leave it unchanged. For example, to three places, 1/3 is rounded to 0.333, and to two places, 0.33; 2/3 is rounded to 0.667 and 0.67.
Other examples: $1.75, rounded to the nearest dime, is $1.80, since the digit dropped is 5; to the nearest dollar, $1.75 rounds to $2. A bill of $12.87 (next page) rounds to $12.90 to the nearest dime, $13 to the nearest dollar, and $10 to the nearest ten dollars. Rounding off just makes clear what we mean when we say $12.87 is "about $13."
where appropriate, ...
If you're figuring how much to tip in a restaurant, you certainly don't need to work it out to the last penny — unless you want to establish a reputation as a pennypincher, even while trying to be generous with a tip. Just an approximate amount is all that's expected. Often you want to give a 15% tip. The easiest way to do this is to first figure 10% by moving the decimal point one place to the left (see pp. 110 and 116), then take half of this, which will be 5%, and add the two. Round off as you proceed. Suppose your bill is $12.87; 10% of this is about $1.30, and half of $1.30 is $0.65. Adding the two, you have $1.95. So if you leave about $2 you'll be doing what you wanted to do.
Many items are priced at just less than an even dollar figure, for sales purposes: $8.99, $13.95, and so on. You don't usually care about the pennies, so just round these up to the nearest dollar: $8.99 + $13.95 is about $9 + $14, or $23.
but be aware that errors can creep in.
It's easier, of course, to work with numbers when you approximate and round off — that's why you do it. But you pay a penalty, as your answers are less accurate. So do it when accuracy is not at a premium. For example, in adding 873 and 465, you might use 900 and 500, getting 1,400. The correct answer of 1,338 is quite a bit different. Does it matter? You must be the judge.
Larger errors can be made in multiplying. For 67 × 31, you might try 70 × 30, or 2,100. Not bad: the correct answer is 2,077. But suppose you simplified 72 × 26 to 70 × 30. The result, 2,100 again, is quite a bit larger than the correct answer, 1,872. There is too much change in the smaller number: 26 became 30. Instead of adding 26 72's together (p. 84), you're adding 30, or four more of the larger number than there should be. The principle: be careful in rounding off numbers before multiplying — particularly the smaller of two numbers.
Ready for problems? Mental exercises help.
Doing math in your head is unfamiliar ground; even with the suggestions given, you may have trouble. One thing you can do to feel more comfortable is to use idle moments to practice. First, a logical extension to the "tables" you learned long ago is to add and multiply twodigit numbers. Start with easier ones like 11,12 and 13 and work up. Try doing it in different ways, using the suggestions in this chapter and perhaps some of your own. Next, become familiar with fraction/decimal equivalents, for fractions with denominators ranging from 2 to 12. Don't try to memorize; look for relations (such as 5/12 = 3/12 + 2/12, which is ¼ + 1/6, or 0.25 + 0.167, or 0.417). Finally, devise problems for yourself, such as converting monthly to annual bills or salaries. You can do 12 × $750 in three ways: (10 + 2) × $750,12 × ($700 + $50), and (¾) × 12 × $1,000. After a while, you'll find it so easy and satisfying you'll wonder why you haven't been using your head more all along.
CHAPTER 2PROBLEMS AND SOLUTIONS
The usual "book" problems don't suit our purpose, ...
Problems can be stated in different ways; shown here are three common types. You'll find many problems of Type 1 in arithmetic books. We've all done these and gained some good practice with numbers. You also find problems of Type 2, where the numbers are not lined up vertically and are a little harder to work with. And then there are "word problems" — Type 3 — which try for realism.
As mentioned in the Introduction, problems that you come across in daily life are not like this. They're certainly different than Types 1 and 2, since you have to decide what to do — there's no +, , × or ÷ sign to tell you. But they also differ from Type 3; real problems often don't even give you numbers, as such, to work with; you have to deal with numbers as words in your mind. Notice that with Type 3, the numbers appear as numbers, not as words. How can problems be stated so they're more like those you'll be doing in your head?
(Continues...)
Excerpted from Doing Simple Math in Your Head by W. J. Howard. Copyright © 1992 W. J. Howard. Excerpted by permission of Chicago Review Press Incorporated.
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by DialABook Inc. solely for the personal use of visitors to this web site.
Table of Contents
Introduction  1  
1  Making Things Easier  7 
2  Problems and Solutions  25 
3  Background: Basic Arithmetic  79 
Glossary  119  
Index  127 
Reading Group Guide
Introduction  1  
1  Making Things Easier  7 
2  Problems and Solutions  25 
3  Background: Basic Arithmetic  79 
Glossary  119  
Index  127 
Interviews
Introduction  1  
1  Making Things Easier  7 
2  Problems and Solutions  25 
3  Background: Basic Arithmetic  79 
Glossary  119  
Index  127 
Recipe
Introduction  1  
1  Making Things Easier  7 
2  Problems and Solutions  25 
3  Background: Basic Arithmetic  79 
Glossary  119  
Index  127 
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Even though I was already advanced in math, this book provided me some awesome shortcuts. But if I think about it the concepts are pretty simple. In fact, I even knew some of the principles they taught me. This is a good buy.

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