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1 Mental Math, Estimation, and Calculators

1 Mental Math, Estimation, and Calculators

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132



Chapter 4 Whole-Number Computation—Mental, Electronic, and Written

numbers are 86 and 14 under addition (since 86 + 14 = 100), 25 and 8 under multiplication (since 25 × 8 = 200), and 600 and 30 under division (since 600 ÷ 30 = 20 ).

In part (a) of Example 4.1, adding 15 to 25 produces a number, namely 40, that is

easy to add to 27. Notice that numbers are compatible with respect to an operation.

For example, 86 and 14 are compatible with respect to addition but not with respect

to multiplication.



a. ( 4 × 13) × 25



Calculate the following mentally using properties and/or compatible numbers.

b. 1710 ữ 9c. 86 ì 15



SOLUTION



Reflection from Research

Recent research shows that computational fluency and number

sense are intimately related.

These tend to develop together

and facilitate the learning of the

other (Griffin, 2003).



a. ( 4 × 13) × 25 = 13 × ( 4 ì 25) = 1300

b. 1710 ữ 9 = (1800 − 90 ) ÷ 9 = (1800 ÷ 9) − (90 ÷ 9) = 200 − 10 = 190

c. 86 × 15 = (86 × 10 ) + (86 × 5) = 860 + 430 = 1290 (Notice that 86 × 5 is half of

86 × 10.)





Compensation The sum 43 + (38 + 17 ) can be viewed as 38 + 60 = 98 using

commutativity, associativity, and the fact that 43 and 17 are compatible numbers.

Finding the answer to 43 + (36 + 19) is not as easy. However, by reformulating the

sum 36 + 19 mentally as 37 + 18, we obtain the sum ( 43 + 37 ) + 18 = 80 + 18 = 98.

This process of reformulating a sum, difference, product, or quotient to one that is

more readily obtained mentally is called compensation. Some specific techniques using

compensation are introduced next.

In the computations of Example 4.1(d), 98 was increased by 2 to 100 and then 59

was decreased by 2 to 57 (a compensation was made) to maintain the same sum. This

technique, additive compensation, is an application of associativity. Similarly, additive

compensation is used when 98 + 59 is rewritten as 97 + 60 or 100 + 57. The problem

47 − 29 can be thought of as 48 − 30 ( = 18). This use of compensation in subtraction is

called the equal additions method since the same number (here 1) is added to both 47

and 29 to maintain the same difference. This compensation is performed to make the

subtraction easier by subtracting 30 from 48. The product 48 × 5 can be found using

multiplicative compensation as follows: 48 × 5 = 24 × 10 = 240 . Here, again, associativity

can be used to justify this method.



Left-to-Right Methods To add 342 and 136, first add the hundreds (300 + 100 ), then



the tens ( 40 + 30 ), and then the ones ( 2 + 6 ), to obtain 478. To add 158 and 279, one can

think as follows: 100 + 200 = 300, 300 + 50 + 70 = 420, 420 + 8 + 9 = 437. Alternatively,

158 + 279 can be found as follows: 158 + 200 = 358, 358 + 70 = 428, 428 + 9 = 437.

Subtraction from left to right can be done in a similar manner. Research has found that

people who are excellent mental calculators utilize this left-to-right method to reduce

memory load, instead of mentally picturing the usual right-to-left written method. The

multiplication problem 3 × 123 can be thought of mentally as 3 × 100 + 3 × 20 + 3 × 3

using distributivity. Also, 4 × 253 can be thought of mentally as 800 + 200 + 12 = 1012 or

as 4 × 250 + 4 × 3 = 1000 + 12 = 1012 .



Multiplying Powers of 10 These special numbers can be multiplied mentally in

either standard or exponential form. For example, 100 × 1000 = 100, 000 ,10 4 × 105 = 109,

20 × 300 = 6000 , and 12, 000 × 110, 000 = 12 × 11 × 107 = 1, 320, 000, 000.

Multiplying by Special Factors Numbers such as 5, 25, and 99 are regarded

as special factors because they are convenient to use mentally. For example,

since 5 = 10 ÷ 2 , we have 38 × 5 = 38 × 10 ÷ 2 = 380 ÷ 2 = 190. Also, since

25 = 100 ữ 4, 36 ì 25 = 3600 ữ 4 = 900 . The product 46 × 99 can be thought of as

46(100 − 1) = 4600 − 46 = 4554 . Also, dividing by 5 can be viewed as dividing by 10,

then multiplying by 2. Thus 460 ÷ 5 = ( 460 ữ 10 ) ì 2 = 46 × 2 = 92.



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From Chapter 3, Lesson 1 “Take Apart Tens to Add” from My Math, Volume 1 Common Core State Standards, Grade 2, copyright © 2013

by McGraw-Hill Education,



133



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Chapter 4 Whole-Number Computation—Mental, Electronic, and Written



Calculate mentally using the indicated method.

a. 197 + 248 using additive compensation

b. 125 × 44 using multiplicative compensation

c. 273 − 139 using the equal additions method

d. 321 + 437 using a left-to-right method

e. 3 × 432 using a left-to-right method

f. 456 × 25 using the multiplying by a special factor method

SOLUTION



a. 197 + 248 = 197 + 3 + 245 = 200 + 245 = 445

b. 125 × 44 = 125 × 4 × 11 = 500 × 11 = 5500

c. 273 − 139 = 274 − 140 = 134

d. 321 + 437 = 758 [Think: (300 + 400 ) + ( 20 + 30 ) + (1 + 7 )]

e. 3 × 432 = 1296 [Think: (3 × 400 ) + (3 × 30 ) + (3 × 2 )]

f. 456 × 25 = 114 × 100 = 11, 400 (Think: 25 × 4 = 100. Thus 456 × 25 = 114 × 4 × 25



= 114 × 100 = 11, 400.)



✔ Check for Understanding: Exercise/Problem Set A #1–7

Use estimation to answer the following two questions:

1) How much blood does the average heart pump in a day?

2) Is $10 enough to buy items with the following prices? $1.49, $0.99, $2.14, $3.23, $1.33, $0.25.

How is the type of estimation used to solve these two problems different?



NCTM Standard

All students should develop and

use strategies to estimate the

results of whole-number computations and to judge the reasonableness of such results.



Reflection from Research

Students can develop a better understanding of number

through activities involving estimation (Leutzinger, Rathmell, &

Urbatsch, 1986).



Children’s Literature



Computational Estimation

The process of estimation takes on various forms. The number of beans in a jar may

be estimated using no mathematics, simply a “guesstimate.” Also, one may estimate

how long a trip will be, based simply on experience. Computational estimation is the

process of finding an approximate answer (an estimate) to a computation, often using

mental math. With the use of calculators becoming more commonplace, computational estimation is an essential skill. Next we consider various types of computational estimation.



Front-End Estimation Three types of front-end estimation will be demonstrated.

Range Estimation Often it is sufficient to know an interval or range—that is, a low

value and a high value—that will contain an answer. The following example shows

how ranges can be obtained in addition and multiplication.



www.wiley.com/college/musser

See “How Much, How Many,

How Far, How Heavy, How Long,

How Tall Is 1000?’’ by Helen

Nolan.



Find a range for answers to these computations by using only

the leading digits.

a.



257          b. 294

+ 576

× 53



SOLUTION



a. Sum

257

+ 576



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Low Estimate

200

+ 500

700



High Estimate

300

+ 600

900



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Section 4.1 Mental Math, Estimation, and Calculators 135

Thus a range for the answer is from 700 to 900. Notice that you have to look at

only the digits having the largest place values (2 + 5 = 7 , or 700) to arrive at the low

estimate, and these digits each increased by one (3 + 6 = 9, or 900) to find the high

estimate.

b. Product



Low Estimate



294

× 53



200

×

50

10,000



High Estimate

300

×



60

18,000



Due to the nature of multiplication, this method gives a wide range, here 10,000 to

18,000. Even so, this method will catch many errors.



Algebraic Reasoning

Estimation is a valuable tool of

algebraic reasoning. It can be

used to determine if a solution to

an equation seems reasonable for

the situation.



One-Column/Two-Column Front-End We can estimate the sum 498 + 251 using

the one-column front-end estimation method as follows: To estimate 498 + 251,

think 400 + 200 = 600 (the estimate). Notice that this is simply the low end of the

range estimate. The one-column front-end estimate always provides low estimates in

addition problems as well as in multiplication problems. In the case of 376 + 53 + 417,

the one-column estimate is 300 + 400 = 700 , since there are no hundreds in 53. The

two-column front-end estimate also provides a low estimate for sums and products.

However, this estimate is closer to the exact answer than one obtained from using

only one column. For example, in the case of 376 + 53 + 417, the two-column frontend estimation method yields 370 + 50 + 410 = 830, which is closer to the exact answer

842 than the 700 obtained using the one-column method.

Front-End with Adjustment This method enhances the one-column front-end estimation method. For example, to find 498 + 251, think 400 + 200 = 600 and 98 + 51 is

about 150. Thus the estimate is 600 + 150 = 750. Unlike one-column or two-column

front-end estimates, this technique may produce either a low estimate or a high estimate, as in this example.

Keep in mind that one estimates to obtain a “rough” answer, so all of the preceding forms of front-end estimation belong in one’s estimation repertoire.



Estimate using the method indicated.

a. 503 × 813 using one-column front-end

b. 1200 × 35 using range estimation

c. 4376 − 1889 using two-column front-end

d. 3257 + 874 using front-end adjustment

SOLUTION



Reflection from Research

To be confident and successful

estimators, children need numerous opportunities to practice

estimation and to learn from

their experiences. Thus, the skills

required by students to develop

a comfort level with holistic, or

range-based, estimation need

to be included throughout a

young child’s education (Onslow,

Adams, Edmunds, Waters,

Chapple, Healey, & Eady, 2005).



c04.indd 135



a. To estimate 503 × 813 using the one-column front-end method, think 500 × 800 =

400,000 . Using words, think “5 hundreds times 8 hundreds is 400,000.”

b. To estimate a range for 1200 × 35, think 1200 × 30 = 36,000 and 1200 × 40 = 48,000 .

Thus a range for the answer is from 36,000 to 48,000. One also could use

1000 × 30 = 30,000 and 2000 × 40 = 80,000. However, this yields a wider range.

c. To estimate 4376 − 1889 using the two-column front-end method, think 4300 −

1800 = 2500 . You also can think 43 − 18 = 25 and then append two zeros after the

25 to obtain 2500.

d. To estimate 3257 + 874 using front-end with adjustment, think 3000, but since

257 + 874 is about 1000, adjust to 4000.





Rounding Rounding is perhaps the best-known computational estimation technique. The purpose of rounding is to replace complicated numbers with simpler

numbers. Here, again, since the objective is to obtain an estimate, any of several

rounding techniques may be used. However, some may be more appropriate than



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Chapter 4 Whole-Number Computation—Mental, Electronic, and Written



Figure 4.1



others, depending on the word problem situation. For example, if you are estimating

how much money to take along on a trip, you would round up to be sure that you had

enough. When calculating the amount of gas needed for a trip, one would round the

miles per gallon estimate down to ensure that there would be enough money for gas.

Unlike the previous estimation techniques, rounding is often applied to an answer as

well as to the individual numbers before a computation is performed.

Several different methods of rounding are illustrated next. What is common, however, is that each method rounds to a particular place. You are asked to formulate

rules for the following methods in the problem set.

Round Up (Down) The number 473 rounded up to the nearest tens place is 480

since 473 is between 470 and 480 and 480 is above 473 (Figure 4.1). The number 473

rounded down to the nearest tens place is 470. Rounding down is also called truncating (truncate means “to cut off”). The number 1276 truncated to the hundreds place

is 1200.

Round a 5 Up The most common rounding technique used in schools is the round

a 5 up method. This method can be motivated using a number line. Suppose that we

wish to round 475 to the nearest ten (Figure 4.2).



Figure 4.2



Since 475 is midway between 470 and 480, we have to make an agreement

concerning whether we round 475 to 470 or to 480. The round a 5 up method always

rounds such numbers up, so 475 rounds to 480. In the case of the numbers 471 to 474,

since they are all nearer 470 than 480, they are rounded to 470 when rounding to the

nearest ten. The numbers 476 to 479 are rounded to 480.

One disadvantage of this method is that estimates obtained when several 5s

are involved tend to be on the high side. For example, the “round a 5 up” to the

nearest ten estimate applied to the addends of the sum 35 + 45 + 55 + 65 yields

40 + 50 + 60 + 70 = 220, which is 20 more than the exact sum 200.

Round to the Nearest Even Rounding to the nearest even can be used to avoid

errors of accumulation in rounding. For example, if 475 + 545 ( = 1020 ) is estimated

by rounding up to the tens place or rounding a 5 up, the answer is 480 + 550 = 1030.

By rounding down, the estimate is 470 + 540 = 1010. Since 475 is between 480 and 470

and the 8 in the tens place is even, and 545 is between 550 and 540 and the 4 is even,

round to the nearest even method yields 480 + 540 = 1020.

Estimate using the indicated method. (The symbol “≈” means

“is approximately.”)

a. Estimate 2173 + 4359 by rounding down to the nearest hundreds place.

b. Estimate 3250 − 1850 by rounding to the nearest even hundreds place.

c. Estimate 575 − 398 by rounding a 5 up to the nearest tens place.

SOLUTION



a. 2173 + 4359 ≈ 2100 + 4300 = 6400

b. 3250 − 1850 ≈ 3200 − 1800 = 1400

c. 575 − 398 ≈ 580 − 400 = 180







Round to Compatible Numbers Another rounding technique can be applied to estimate products such as 26 × 37. A reasonable estimate of 26 × 37 is 25 × 40 = 1000. The

numbers 25 and 40 were selected since they are estimates of 26 and 37, respectively,

and are compatible with respect to multiplication. (Notice that the rounding up technique would have yielded the considerably higher estimate of 30 × 40 = 1200, whereas

the exact answer is 962.) This round to compatible numbers technique allows one to

round either up or down to compatible numbers to simplify calculations, rather



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Section 4.1 Mental Math, Estimation, and Calculators 137

than rounding to specified places. For example, a reasonable estimate of 57 × 98

is 57 × 100 ( = 5700 ). Here, only the 98 needed to be rounded to obtain an estimate

mentally. The division problem 2716 ÷ 75 can be estimated mentally by considering

2800 ÷ 70 ( = 40 ). Here 2716 was rounded up to 2800 and 75 was rounded down to 70

because 2800 ÷ 70 easily leads to a quotient since 2800 and 70 are compatible numbers

with respect to division.

Reflection from Research

Good estimators use three computational processes. They make

the number easier to manage

(possibly by rounding), change

the structure of the problem itself

to make it easier to carry out, and

compensate by making adjustments in their estimation after

the problem is carried out (Reys,

Rybolt, Bestgen, & Wyatt, 1982).



a. 43 × 21



Estimate by rounding to compatible numbers in two different

ways.

b. 256 ữ 33



SOLUTION



a. 43 ì 21 40 × 21 = 840

43 × 21 ≈ 43 × 20 = 860

(The exact answer is 903.)



b. 256 ÷ 33 ≈ 240 ÷ 30 = 8

256 ÷ 33 ≈ 280 ÷ 40 = 7

(The exact answer is 7 with remainder 25.)







Rounding is a most useful and flexible technique. It is important to realize that

the main reasons to round are (1) to simplify calculations while obtaining reasonable

answers and (2) to report numerical results that can be easily understood. Any of the

methods illustrated here may be used to estimate.

The ideas involving mental math and estimation in this section were observed in

children who were facile in working with numbers. The following suggestions should

help develop number sense in all children:

1. Learn the basic facts using thinking strategies, and extend the strategies to multidigit numbers.

2. Master the concept of place value.

3. Master the basic addition and multiplication properties of whole numbers.

4. Develop a habit of using the front-end and left-to-right methods.

5. Practice mental calculations often, daily if possible.

6. Accept approximate answers when exact answers are not needed.

7. Estimate prior to doing exact computations.

8. Be flexible by using a variety of mental math and estimation techniques.



✔ Check for Understanding: Exercise/Problem Set A #8–15



Vincent LaRussa/John Wiley and Sons



Using a Calculator



Figure 4.3



c04.indd 137



Although a basic calculator that costs less than $10 is sufficient for most elementary

school students, there are features on $15 to $30 calculators that simplify many

complicated calculations. Also, most smart phones have basic as well as scientific

calculators. When we use the word calculator throughout the book, it may encompass many of these types of calculators. The TI-34 MultiView, manufactured by

Texas Instruments, is shown in Figure 4.3. The TI-34 MultiView, which is designed

especially for elementary and middle schools, performs fraction as well as the usual

decimal calculations, can perform long division with remainders directly, and has the

functions of a scientific calculator. One nice feature of the TI-34 MultiView is that it

has mutiple lines of display, which allows the student to see the input and output at

the same time.

The on key turns the calculator on. The delete key is an abbreviation for “delete”

and allows the user to delete one character at a time from the right if the cursor is at

the end of an expression or delete the character under the cursor. Pressing the clear

key will clear the current entry. The previous entry can be retrieved by pressing the

m on the upper right key.

Two types of logic are available on common calculators: arithmetic and algebraic.



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138



Chapter 4 Whole-Number Computation—Mental, Electronic, and Written



Reflection from Research

Students in classrooms where

calculators are used tend to have

more positive attitudes about

mathematics than students in

classrooms where calculators are

not used (Reys & Reys, 1987).



Common Core –

Mathematical Practices

Use appropriate tools strategically.

Proficient students are sufficiently

familiar with tools appropriate

for their grade or course to make

sound decisions about when each

of these tools might be helpful,

recognizing both the insight to be

gained and their limitations.



(NOTE: For ease of reading, we will write numerals without the usual squares

around them to indicate that they are keys.)



Arithmetic Logic In arithmetic logic, the calculator performs operations in the

order they are entered. For example, if 3 + 4 × 5 = is entered, the calculations are

performed as follows: (3 + 4 ) × 5 = 7 × 5 = 35. That is, the operations are performed

from left to right as they are entered.

Algebraic Logic If your calculator has algebraic logic and the expression 3 +

4 × 5 = is entered, the result is different; here the calculator evaluates expressions

according to the usual mathematical convention for order of operations.

If a calculator has parentheses, they can be inserted to be sure that the desired

operation is performed first. In a calculator using algebraic logic, the calculation

13 5 ì 4 ữ 2 + 7 will result in 13 − 10 + 7 = 10 . If one wishes to calculate 13 − 5 first,

parentheses must be inserted. Thus (13 − 5) × 4 ÷ 2 + 7 = 23.

Use the order of operations to mentally calculate the following

and then enter them into a calculator to compare results.



Reflection from Research

Young people in the workplace

believe that menial tasks, such as

calculations, are low-order tasks

that should be undertaken by

technology and that their role is

to identify problems and solve

them using technology to support that solution. Thus, mathematics education may need

to reshift its focus from accuracy

and precision relying on arduous

calculations to one that will better fit the contemporary workplace and life beyond schools

(Zevenbergen, 2004).



a. ( 4 + 2 × 5) ÷ 7 + 3

b. 8 ÷ 22 + 3 × 22

c. 17 − 4(5 − 2 )

d. 40 ÷ 5 × 23 − 2 × 3

SOLUTION



a. ( 4 + 2 ì 5) ữ 7 + 3 = ( 4 + 10 ) ÷ 7 + 3

= (14 ÷ 7 ) + 3

=2+3=5

b. 8 ữ 22 + 3 ì 22 = (8 ữ 4 ) + (3 ì 4 )

= 2 + 12 = 14

c. 17 − 4(5 − 2 ) = 17 − ( 4 × 3)

= 17 − 12 = 5

d. 40 ữ 5 ì 23 2 × 3 = [( 40 ÷ 5) × 23 ] − ( 2 × 3)

=8×8−2×3

= 64 − 6 = 58







Now let’s consider some features that make a calculator helpful both as a computational and a pedagogical device. Several keystroke sequences will be displayed to

simulate the variety of calculator operating systems available.



Parentheses As mentioned earlier when we were discussing algebraic logic, one

must always be attentive to the order of operations when several operations are present. For example, the product 2 × (3 + 4 ) can be found in two ways. First, by using

commutativity, the following keystrokes will yield the correct answer:

3 + 4 = × 2 =



14 .



Alternatively, the parentheses keys may be used as follows:

2 × ( 3 + 4 ) =



14 .



Parentheses are needed, since pressing the keys 2 × 3 + 4 = on a calculator with

algebraic logic will result in the answer 10. Distributivity may be used to simplify

calculations. For example, 753 ¥ 8 + 753 ¥ 9 can be found using 753 × ( 8 + 9 ) =

instead of 753 × 8 + 753 × 9 = . The enter key on the TI-34 MultiView is often

used in place of an equals key.



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Section 4.1 Mental Math, Estimation, and Calculators 139



Constant Functions In Chapter 3, multiplication was viewed as repeated addi-



tion; in particular, 5 × 3 = 3 + 3 + 3 + 3 + 3 = 15. Repeated operations are carried out

in different ways depending on the model of calculator. For example, the following

keystroke sequence is used to calculate 5 × 3 on one calculator that has a built-in

constant function:

3 + = = = =

15 .



Numbers raised to a whole-number power can be found using a similar technique.

For example, 35 can be calculated by replacing the + with a × in the preceding

examples. A constant function can also be used to do repeated subtraction to find a

quotient and a remainder in a division problem. For example, the following sequence

can be used to find 35 ÷ 8:

35 − 8 = = = =



3.



The remainder (3 here) is the first number displayed that is less than the divisor (8

here), and the number of times the equal sign was pressed is the quotient (4 here).

Because of the multiple lines of display, the TI-34 MultiView can handle most of

these examples by typing in the entire expression. For example, representing 5 × 3 as

repeated addition is entered into the TI-34 MultiView as

3 + 3 + 3 + 3 + 3 enter

The entire expression of 3 + 3 + 3 + 3 + 3 appears on the first line of display and the

15 appears on the end of the same line of display when mode is NORM.

result



Exponent Keys There are three common types of exponent keys: x 2 , y2 , and ∧ .

The x 2 key is used to find squares in one of two ways:

3 x2 =



9



or 3 x 2



9.



The y and ∧ keys are used to find more general powers and have similar keystrokes. For example, 73 may be found as follows:

x



7 yx 3 =



343



or 7 ∧ 3



343 .



Memory Functions Many calculators have a memory function designated by

the keys M+ , M − , MR , or STO , RCL , SUM . Your calculator’s display will

probably show an “M” to remind you that there is a nonzero number in the memory.

The problem 5 × 9 + 7 × 8 may be found as follows using the memory keys:

5 × 9 = M+ 7 × 8 = M+ MR



101



5 × 9 = SUM 7 × 8 = SUM RCL



or

101 .



It is a good practice to clear the memory for each new problem using the all clear key.

The TI-34 MultiView has seven memory locations—x, y, z, t, a, b, c—which can be

yzt

accessed by pressing the sto c key and then pressing the x abc

key multiple times to

select the desired variable. The following keystrokes are used to evaluate the expression above.

5 × 9 sto c



x



yzt

abc



enter



and 7 × 8 sto c



x



yzt

yzt

x

abc

abc



enter



The above keys will store the value 45 in the memory location x and the value 56 in the

memory location y. The values x and y can then be added together as follows.

2nd recall enter + 2nd recall . enter enter

This will show 45 + 56 on the left of the calculator display and 101 on the right of the

display.

Additional special keys will be introduced throughout the book as the need arises.



Scientific Notation Input and output of a calculator are limited by the number

of places in the display (generally 8, 10, or 12). Two basic responses are given when



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Chapter 4 Whole-Number Computation—Mental, Electronic, and Written

a number is too large to fit in the display. Simple calculators either provide a partial answer with an “E” (for “error”), or the word “ERROR” is displayed. Many

scientific calculators automatically express the answer in scientific notation (that is,

as the product of a decimal number greater than or equal to 1 but less than 10, and

the appropriate power of 10). For example, on the TI-34 MultiView, the product

of 123,456,789 and 987 is displayed 1.218518507 × 1011 (NOTE: Scientific notation

is discussed in more detail in Chapter 9 after decimals and negative numbers have

been studied.) If an exact answer is needed, the use of the calculator can be combined with paper and pencil and distributivity as follows:

123, 456, 789 × 987 = 123, 456, 789 × 900 + 123, 456, 789

× 80 + 123, 456, 789 × 7

= 111,111,110,100 + 9, 876, 543,120

+ 864,197, 523.

Now we can obtain the product by adding:

111,111,110,100

9, 876, 543,120

+

864,197, 523

121, 851, 850, 743

Calculations with numbers having three or more digits will probably be performed

on a calculator (or computer) to save time and increase accuracy. Even so, it is prudent to estimate your answer when using your calculator.



Shakuntala Devi (1929-2013), an Indian mathematical wizard, was dubbed

‘the human computer’ because she could calculate swiftly mentally. At six,

she performed at circuses and other road shows and became the source

of support for her family. In 1977, she found the 23rd root of a 201-digit

number in 50 seconds while a Univac computer required 62 seconds. In

1982, she earned a place in Guinness Book of World Records for mentally

multiplying two 13 digit numbers in 28 seconds. She also had careers in

astrology and as a cookbook author and a novelist.



©Ron Bagwell



✔ Check for Understanding: Exercise/Problem Set A #16–24



EXERCISE/PROBLEM SET A

EXERCISES



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1. Calculate mentally using properties.

a. (37 + 25) + 43        b. 47 ¥ 15 + 47 Ơ 85

c. ( 4 ì 13) ì 25d. 26 ¥ 24 − 21 ¥ 24



3. Calculate mentally left to right.

a. 123 + 456

b. 342 + 561

c. 587 − 372

d. 467 − 134



2. Find each of these differences mentally using equal

additions. Write out the steps that you thought

through.

a. 43 − 17                  b. 62 − 39

c. 132 − 96                 d. 250 − 167



4. Calculate mentally using the indicated method.

a. 198 + 387 (additive compensation)

b. 84 × 5 (multiplicative compensation)

c. 99 × 53 (special factor)

d. 4125 ÷ 25 (special factor)



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Section 4.1 Mental Math, Estimation, and Calculators 141

5. Calculate mentally.

a. 58,000 × 5,000,000

,

b. 7 × 105 × 21000

c. 13,000 × 7,000,000

d. 4 × 105 × 3 × 106 × 7 × 103

e. 5 × 103 × 7 × 107 × 4 × 105

f. 17,000,000 × 6,000,000,000



14. Here are four ways to estimate 26 × 12:



6. The sum 26 + 38 + 55 can be found mentally as follows:

26 + 30 = 56, 56 + 8 = 64, 64 + 50 = 114, 114 + 5 = 119. Find

the following sums mentally using this technique.

a. 32 + 29 + 56

b. 54 + 28 + 67

c. 19 + 66 + 49

d. 62 + 84 + 27 + 81



15. Estimate the following values and check with a calculator.

a. 656 × 74 is between _____ 000 and _____ 000.

b. 491 × 3172 is between _____ 00000 and _____ 00000.

c. 1432 is between _____ 0000 and _____ 0000.



26 × 10 = 260 30 × 12 = 360

25 × 12 = 300 30 × 10 = 300

Estimate the following in four ways.

a. 31 × 23

b. 35 × 46

c. 48 × 27

d. 76 × 12



7. In division you can sometimes simplify the problem by

multiplying or dividing both the divisor and dividend by

the same number. This is called division compensation. For

example,

72 ÷ 12 = (72 ÷ 2 ) ÷ (12 ÷ 2 ) = 36 ÷ 6 = 6 and

145 ÷ 5 = (145 × 2 ) ÷ (5 × 2 ) = 290 ÷ 10 = 29

Calculate the following mentally using this technique.

a. 84 ÷ 14       b. 234 ÷ 26       c. 120 ÷ 15      d. 168 ÷ 14

8. Before granting an operating license, a scientist has to

estimate the amount of pollutants that should be allowed

to be discharged from an industrial chimney. Should she

overestimate or underestimate? Explain.

9. Estimate each of the following using the four front-end

methods: (i) range, (ii) one-column, (iii) two-column, and

(iv) with adjustment.

a.



3741

+ 1252



b. 1591

346

589

+ 163



c. 2347

58

192

+ 5783



16. Guess what whole numbers can be used to fill in the

blanks. Use your calculator to check.

a. _____ 6 = 4096

b. _____ 4 = 28,561

17. Guess which is larger. Check with your calculator.

a. 54 or 45 ?

b. 73 or 37 ?

c. 7 4 or 47 ?



d. 63 or 36 ?



18. Some products can be found most easily using a

combination of mental math and a calculator. For

example, the product 20 × 47 × 139 × 5 can be found by

calculating 47 × 139 on a calculator and then multiplying

your result by 100 mentally ( 20 × 5 = 100 ). Calculate the

following using a combination of

mental math and a calculator.

a. 17 × 25 × 817 × 4

b. 98 × 2 × 673 × 5

c. 674 × 50 × 889 × 4

d. 783 × 8 × 79 × 125

19. Compute the quotient and remainder (a whole number)

for the following problems on a calculator without using

repeated subtraction. Describe the procedure you used.



)



)



a. 8 103



b. 17 543



)



c. 123 849

10. Find a range estimate for these products.

a. 37 × 24        b. 157 × 231        c. 491 × 8

11. Estimate using compatible number estimation.

a. 63 ì 97

b. 51 ì 212

c. 3112 ữ 62

d. 103 ì 87

e. 62 ì 58

f. 4254 ữ 68

12. Round as specified.

a. 373 to the nearest tens place

b. 650 using round a 5 up method to the hundreds place

c. 1123 up to the tens place

d. 457 to the nearest tens place

e. 3457 to the nearest thousands place

13. Cluster estimation is used to estimate sums and products

when several numbers cluster near one number. For

example, the addends in 789 + 810 + 792 cluster around

800. Thus 3 × 800 = 2400 is a good estimate of the sum.

Estimate the following using cluster estimation.

a. 347 + 362 + 354 + 336

b. 61 × 62 × 58

c. 489 × 475 × 523 × 498

d. 782 + 791 + 834 + 812 + 777



c04.indd 141



)



d. 894 107, 214



20. 1233 = 12 + 33 and 8833 = 882 + 332. How about 10,100

and 5,882,353? (Hint: Think 588 2353.)

2



2



21. Determine whether the following equation is true for

n = 1, 2, or 3.

1n + 6 n + 8n = 2 n + 4 n + 9n

22. Notice that 153 = 13 + 53 + 33. Determine which of the following numbers have the same property.

a. 370

b. 371

c. 407

23. Determine whether the following equation is true when n

is 1, 2, or 3.

1n + 4 n + 5n + 5n + 6 n + 9n =

2 n + 3n + 3n + 7 n + 7 n + 8n

24. Simplify the following expressions using a calculator.

a. 135 − 7( 48 − 33)

b. 32 ¥ 512

c. 132 ¥ 34 − 3( 45 − 37 )3

26 + 4(313 − 172 2 )2 + 2 ¥ 401

d.

26 ¥ (313 − 172 2 )



7/30/2013 2:52:56 PM



142



Chapter 4 Whole-Number Computation—Mental, Electronic, and Written



PROBLEMS

The answer is 71 consecutive 1s—one of the biggest numbers a computer has ever factored. This factorization

bested the previous high, the factorization of a 69-digit

number. Find a mistake here, and suggest a correction.



25. Five of the following six numbers were rounded to

the nearest thousand, then added to produce an estimated

sum of 87,000. Which number was not included?

5228 14, 286 7782 19, 628 9168 39, 228

26. True or false?

493, 827,156 2 = 246, 913, 578 × 987, 654, 312

27. Place a multiplication sign or signs so that the product in

each problem is correct; for example, in 1 2 3 4 5 6 =

41,472, the multiplication sign should be between the

2 and the 3 since 12 × 3456 = 41, 472.

a. 1 3 5 7 9 0 = 122,130

b. 6 6 6 6 6 6 = 439,956

c. 7 8 9 3 4 5 6 = 3,307,824

d. 1 2 3 4 5 6 7 = 370,845



34. Some mental calculators use the following fact:

( a + b )( a − b ) = a 2 − b 2. For example, 43 × 37 =

( 40 + 3)( 40 − 3) = 40 2 − 32 = 1600 − 9 = 1591. Apply this

technique to find the following products mentally.

a. 54 × 46

b. 81 × 79

c. 122 × 118

d. 1210 × 1190

35. Fermat claimed that

100, 895, 598,169 = 898, 423 × 112, 303.

Check this on your calculator.



28. Develop a method for finding a range for subtraction

problems for three-digit numbers.



36. Show how to find 439, 268 × 6852 using a calculator that

displays only eight digits.



29. Find 13, 333, 3332.



37. Insert parentheses (if necessary) to obtain the following

results.

a. 76 × 54 + 97 = 11, 476

b. 4 × 132 = 2704

c. 13 + 592 × 47 = 163, 620

d. 79 − 43 ÷ 2 + 17 2 = 307



30. Calculate 99 ¥ 36 and 99 ¥ 23 and look for a pattern. Then

predict 99 ¥ 57 and 99 ¥ 63 mentally and check your predictions with a calculator.

31. a. Calculate 252 , 352 , 452 , and 552 and look for a

pattern. Then find 652 , 752 , and 952 mentally and

check your answers.

b. Using a variable, prove that your result holds for squaring numbers that have a 5 as their ones digit.

32. George Bidder was a calculating prodigy in England during the nineteenth century. As a nine-year-old, he was

asked: If the moon were 123,256 miles from the Earth and

sound traveled at the rate of 4 miles a minute, how long

would it be before inhabitants of the moon could hear

the battle of Waterloo? His answer—21 days, 9 hours, 34

minutes—was given in 1 minute. Was he correct? Try to

do this calculation in less than 1 minute using a calculator.

(NOTE: The moon is about 240,000 miles from Earth and

sound travels about 12.5 miles per second.)

33. Found in a newspaper article: What is

241, 573, 142, 393, 627, 673, 576, 957, 439, 048 × 45, 994,

811, 347, 886, 846, 310, 221, 728, 895, 223, 034, 301, 839?



38. a. Find a shortcut.

24 × 26 = 624

62 × 68 = 4216

73 × 77 = 5621

41 × 49 = 2009

86 × 84 = 7224

57 × 53 =

b. Prove that your result works in general.

c. How is this problem related to Problem 31?

39. Develop a set of rules for the round a 5 up method.

40. There are eight consecutive odd numbers that when multiplied together yield 34,459,425. What are they?

41. Jill goes to get some water. She has a 5-liter pail and a

3-liter pail and is supposed to bring exactly 1 liter back.

How can she do this?



EXERCISE/PROBLEM SET B

EXERCISES



c04.indd 142



1. Calculate mentally using properties.

a. 52 ¥ 14 − 52 ¥ 4

b. (5 × 37 ) × 20

c. (56 + 37 ) + 44

d. 23 ¥ 4 + 23 ¥ 5 + 7 ¥ 9



3. Calculate mentally using the left-to-right method.

a. 246 + 352

b. 49 + 252

c. 842 − 521

d. 751 − 647



2. Find each of these differences mentally using

equal additions. Write out the steps that you thought

through.

a. 56 − 29

b. 83 − 37

c. 214 − 86

d. 542 − 279



4. Calculate mentally using the method indicated.

a. 359 + 596 (additive compensation)

b. 76 × 25 (multiplicative compensation)

c. 37 ì 98 (special factor)

d. 1240 ữ 5 (special factor)



7/30/2013 2:53:01 PM



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