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 easy-to-use 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 forbidding-looking 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...