Mathematical Discourse || Whole Number Multiplication: There's More to It Than Might Be Expected!

Whole Number Multiplication: There's More to It Than Might Be Expected!Author(s): Christy D. Graybeal, Al Cuoco and E. Paul GoldenbergSource: The Mathematics Teacher, Vol. 101, No. 4, Mathematical Discourse (NOVEMBER 2007),pp. 312-316Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/20876117 .

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Christy D. Graybeal

Whole Number Multiplication:

There's More to It Than

Might Be Expected!

The problem: Arrange the digits 5, 6, 7, 8, and 9 into a three-digit number and a two-digit num

ber so that.their product is as large as possible.

efore you read any further, write down

your prediction. Did you guess 987 x 65 or

maybe 975 x 86? Do not feel embarrassed? those were my first guesses as well! Although it is not immediately obvious, there is more to this prob lem than might at first be expected.

MOTIVATION My interest in this problem arose while I was tutor

ing a seventh grader. We were flipping through the book Math Mind Stretchers (Fisher 1997) and came across this and similar problems. My student began by making educated guesses and finding the prod ucts. Just as I would have, he placed the largest digit (9) in the hundreds place. As he worked though

different possibilities, I quickly realized that I did not know what the largest product was or even an

elegant way of determining it. My approach to the

problem would have been very similar to my stu dent's. I began thinking that there must be a more

elegant method.

Further, I was surprised by the answer given in the back of the book: 875 x 96. As is often the case in mathematics, the answer generated more ques tions: Why is it better to have the largest digit in the tens place of the second number instead of in the hundreds place of the first number? What is the

reasoning behind the placement of the digits? In the

spirit of Brown and Walter (1993), I used this prob lem to pose new questions as well: What if the digits are not consecutive? What if all the digits are not dis tinct? Does this pattern hold for any five digits? The

more I explored these questions, the more I learned about whole-number multiplication. At the same

time, I also learned about the importance of repre sentations. Different representations of this problem led me to different insights and understandings.

SOLUTION STRATEGIES Because I was working with a prealgebra middle school student when I became interested in this

problem, I initially tried to solve the problem as

my student did, using the guess-and-check method. I selected five distinct digits and tried different

arrangements until I found what I believed was the

largest product. The difficulty with this approach is knowing whether I had in fact obtained the larg est product. Several times I had what I thought was a winner, only to find a better arrangement later.

Without trying all 120 possible arrangements of five distinct digits, could I guarantee that the prod uct I obtained was the largest possible?

This department focuses on mathematics content that appeals to secondary school

teachers. It provides a forum that allows classroom teachers to share their mathemat

ics from their work with students, their classroom investigations and projects, and their other experiences. We encourage submissions that pose and solve a novel or interest

ing mathematics problem, expand on connections among different mathematical topics,

present a general method for describing a mathematical notion or solving a class of prob

lems, elaborate on new insights into familiar secondary school mathematics, or leave the

reader with a mathematical idea to expand. Send submissions to "Delving Deeper" by

accessing mt.msubmit.net.

"Delving Deeper" can accept manuscripts in ASCII or Word formats only.

Edited by Al Cuoco, [email protected] Center for Mathematics Education, Education Development Center

Newton, MA 02458

E. Paul Goldenberg, [email protected] Center for Mathematics Education, Education Development Center

Newton, MA 02458

312 MATHEMATICS TEACHER | Vol. 101, No. 4 November 2007



2nd 3rd 5th

largest largest largest

largest 4th

largest

Fig. 1 This arrangement seemed to produce the largest

product.

dor e

dor e

Fig. 2 The largest digits must be placed as shown.

d or e b or c

dor e b or c

Fig. 3 The next two largest digits must be placed as shown.

As I experimented with various sets of five digits, a pattern emerged. For five digits, it seemed that the

arrangement in figure 1 resulted in the largest prod uct. With conjecture in hand, I set out to prove this.

Along the way I gained new insights into the multipli cation algorithm and multiplication itself. By examin

ing the algorithm, I noticed how different arrange ments affect the product. Switching the position of two digits results in gains in some places and losses in

others; by maximizing gains and minimizing losses, one can increase the product. Early in my work I realized why my initial guesses did not result in the

largest product; later, I found that the arrangement in

figure 1 does not always result in the largest product. Two representations of the problem (informal and

formal) are presented, each offering different insights and each having its own benefits and drawbacks.

The informal approach helped me see what was

happening in the multiplication algorithm and why my conjectured arrangement usually resulted in the

largest product. However, it did not satisfy my intrin sic need for formal proof, nor did it, at first, help me see when or why there is an exception to the arrange

ment. (In fact, I did not realize that there was an

exception until I considered the more formal proof.) The more formal approach satisfies my need for

proof, and makes the special case more obvious but does not provide as much insight into the multiplica tion algorithm. The symbol manipulations, at times,

mask the quantities involved. For me, this proof is what Hanna (1989) would consider a "proof that

proves" rather than a "proof that explains."

INFORMAL APPROACH For five distinct digits, there are 5! = 120 possible arrangements. But it is already clear, even to a young student, that one does not need to investigate all 120

possibilities. Common sense (more precisely, place value and even a beginner's understanding of the

multiplication algorithm) eliminates many of these. Visual inspection also helps suggest what compari sons need to be made to refine the initial guesses.

Although I worked in this approach informally, writ

ing it up is easiest with formal symbolism.

Consider the digits a, b, c, d, and e where 0 < a <

b<c<d<e<9.

Although I originally thought of these digits as distinct, one of my questions was about what would happen if some digits repeated. Thus, I did not use a strict inequality in the problem setup.

To solve the initial problem, we want to maxi mize the thousands, then the hundreds, and so on. To maximize the number of thousands, the two

largest digits, d and e, should be in the leftmost

positions of both the three-digit number and the

two-digit number (see fig. 2). To maximize the number of hundreds, the next two largest digits, b and c, should be in the tens place of the three-digit number and the ones place of the two-digit number

(see fig. 3). By default, the smallest digit, a, should be in the

ones place of the three-digit number. This results in four possibilities (see fig. 4).

In summary:

Case 1: (lOOd + 101? + a){l0e + c) = lOOOd* +

100{be + cd) + \0{ae + be) + ac Case 2: (lOOd + 10c + a)(10c + b) = lOOOdc +

100{bd + ce) + 10[ae + be) + ab

Vol. 101, No. 4 November 2007 | MATHEMATICS TEACHER 313



Case 1

Case 3

Fig. 4 After placing a, there are four cases.

Case 2

c

e

Case 4

c

Case 3: (100* + 101? + a) (lOd + *) = lOOOd* +

100(bd + **) + \0{ad + I?*) + ac Case 4: (100* + 10* + a)(10d + b) = 1000d* +

100(1?* + cd) + \0{ad + &*) + ab

All four cases have the same number (de) repre senting the thousands.

To compare the number of hundreds, I needed to determine whether (bd + **) or {be + cd) is

greater.

Given d<e,b<c^>

(* -

b)d < (* -

b)e^ cd -bd< ce - be ̂

be + cd<bd + ce. So Case 2 and Case 3 have at least as many hun dreds as Case 1 and Case 4.

To compare the number of tens, I needed to determine whether {ad + be) or [ae + be) is greater.

Given d < e, ad<ae->

ad + bc<ae + be.

So Case 1 and Case 2 have at least as many tens as Case 3 and Case 4.

To compare the number of ones, I needed to determine whether ab or ac is greater.

Given b < c implies ab < ac. So Case 1 and Case 3 have at least as many ones as Case 2 and Case 4.

In summary:

All four arrangements have the same number of thousands.

Case 2 and Case 3 have at least as many hundreds as Case 1 and Case 4. Case 1 and Case 2 have at least as many tens as Case 3 and Case 4. Case 1 and Case 3 have at least as many ones as Case 2 and Case 4.

At first glance it seems that since Case 2 has the greatest number of hundreds and tens, it will

always result in the largest product. The exception to this did not become obvious until I developed a

more formal proof.

FORMAL PROOF To determine with greater certainty which of these four cases results in the largest product, we will

compare them in pairs.

Comparison of Case 1 with Case 2 Subtract Case 1 from Case 2 and consider the sign of the difference. Case 2 - Case 1:




[lOOOdc + 100(fod + ce) + lO(ae + be) + ab] - [lOOOde + 100(foe + cd) + 10(W+ foe) + ac\

= 100(bd + cc) -

100(foe + cd) + ab - ac = 100(fod + ce - foe - cd) + a(fo

- c)

= 100(b - c)(d - e) + a(b - c) = (fo-c)[100(d-e) + a]

The following lemma will be useful in determining the sign of the difference.

Lemma 1. If d < e, then [I00(d -e)+a] is negative. If d = e, then [100(d -e)+a] is

nonnegative.

Proof. If d < e, then d - e < -1, and 100(d -

e) <

-100, which makes [I00(d -

e) + a] negative. If d = e, then 0 = 100(d

- e) and [100(d

- e) + a] is

nonnegative.

Since fo < c, (fo -

c) is nonpositive. If d < e, then Case 2 - Case 1 is nonnegative and Case 1 < Case 2. If d = e, then Case 2 - Case 1 is nonpositive and Case 1 > Case 2.

Comparison of Case 2 to Cases 3 and 4 follows a

similar plan.

Comparison of Case 2 with Case 3 Subtract Case 3 from Case 2 and consider the sign of the difference. Case 2 - Case 3:

[lOOOde + 100(fod + ce) + 10(oe + be) + ab] - [lOOOde + 100(fod + ce) + 10(ad + be) + ac]

= 10(ae + be) -

10(ad + be) +ab- ac = 10(ae + bc-ad- be) + a(b -

c) = lOGze

- ad) + a(fo

- c)

= a[\0{e - d) + (fo - c)]

Lemma 2. If d < e, then [10(e -d) + (b- c)] is

positive.

Tfd = e, then [I0(e - d) + (fo - c)] is nonpositive.

Proof. If d < e, then e - d > 1 and 10(e -

d) > 10. Since 0<fo<c<9, -9<fo-c<0. Thus, [I0(e

- d) +

(fo -

c)] is positive. If d = e, then 10(e -

d) = 0. Since fo < c, fo - c is nonpositive and [10(e

- d) + (fo

- c)] is

nonpositive.

If d < e, then Case 2 - Case 3 is nonnegative and Case 3 < Case 2. If d = e, then Case 2 - Case 3 is nonpositive and Case 3 > Case 2.

Comparison of Case 2 with Case 4

Again, we subtract Case 4 from Case 2 and consider the sign of the difference. Case 2 - Case 4:

[lOOOde + 100(bd + ce) + 10(W + be) + ab] - [lOOOde? + 100(be + cd) + 10(ad + fc) + afc] = 100(fe2 + ce) - 100(fe + cd) + 10(a<? + for) -

\0{ad + be) = 100(bd + ce-be- cd) + 10(ae

- ad)

= 100(c -

b)[e -

d) + 10a(^ -

d) = 10(e-d)[10(c-b)+a]

Lemma 3. [I0(c -

b) + a] is nonnegative.

Proof. Given b<c,we have 0 < c - b. So 0 <

10(c -

b). Because a is not negative, [I0(c -

b) + a] is nonnegative.

If d < e, then (^ -

d) is positive. Thus, 10(e -

d) [10(c -

b) + a] is nonnegative and Case 4 < Case 2. If d = ?, then Case 2 = Case 4.

Summary of the three comparisons \td<e, then Case 1 < Case 2, Case 3 < Case 2, and Case 4 < Case 2. Thus, Case 2 results in the largest product.

If d - e, then Case 2 < Case 1, Case 2 < Case 3, and Case 2 = Case 4. But d = e also makes Cases 1 and 3 indistinguishable. Thus, Case 1 (Case 3) results in the largest product.

An example shows the significance of this result.

Example Consider the digits 1, 3, 6, 7, and 9. Since the two

largest digits are not equal, Case 2 will result in the

largest product:

761 x 93 = 70,773

Now if the 7 is changed to another 9, one might think that the largest product would be

961 x 93 = 89,373.

However, Case 1 = Case 3 results in a larger product:

931 x96 = 89,376

Originally, this surprised me. How could 3 more

(96 - 93 = 3) of 30 less (961

- 931 = 30) result in a larger product? My informal approach helped me see why.

Call 961 x 93 = 89,373 Case 2. Call 931 x 96 = 89,376 Case 1. Both cases have 9 x 9 = 81 thousands. Both have 6x9 + 3x9 = 81 hundreds. Both have 1x9 + 3x6 = 27 tens.

The difference is in the number of ones: Case 1 has 1x6 = 6 ones. Case 2 has 1x3 = 3 ones.

Thus, Case 1 results in a larger product than Case 2.

Vol. 101, No. 4 November 2007 | MATHEMATICS TEACHER 315



CONCLUDING THOUGHTS My favorite mathematics problems are ones that are deceptively simple?easy to describe and understand yet complex. The generalization of this

problem to any five digits fits this description. As I worked on this problem, I talked to mathemati

cians, mathematics educators, and mathematics

teachers. All have been intrigued by it. Several have asked me how I think the author of the prob lem expected middle school students to solve the

original problem and what they would get out of it. Although I doubt that a middle school student

would approach the problem as I did, I think the

problem is still a good one. For one thing, by considering place value and

the multiplication algorithm, students can make educated guesses about the placement of the digits. Through logic, students will realize that they do not need to check all 120 possibilities. Moreover, stu

dents can gain insights from the informal and for mal approaches and appreciate the value in looking at a single problem in multiple ways. Generalizing the problem to any five digits provides a good les son in the importance of paying attention to detail and being open to surprises. Without these disposi tions, it is easy to overlook the exception of what

happens when the two largest digits are equal. I encourage teachers to help their students

explore this and similar problems. For example, students can explore what happens if the goal is to obtain the smallest product. (The pattern is slightly different.) How can seven digits be arranged into a four-digit number and a three-digit number so

that their product is as large as possible? Is the pat tern similar to the pattern for five digits? Are there

any surprises? By extending the problem, students learn an important lesson in mathematics: There is

always more to learn and explore.

Editors' note: At first, the author's suggestion to

explore the pattern for seven digits caught us by surprise. Why not ask first about six digits? How

much more are we learning as we ask for more

and more digits? Even after we solve the problem for seven digits, we still do not know what might happen with nine?or thirty-three, for that mat ter! Moreover, with seven digits, do we want to restrict ourselves to thinking about the product of a three-digit number multiplied by a four-digit one, or might a still greater (or smaller) product be achieved through a different use of those seven

digits? With seven digits, we might, instead, con struct the product of three numbers?one three

digit number and two two-digit numbers?and that might produce the greatest product. We were about to lose interest in this initially very intrigu ing problem because of the apparent endlessness

of the work. Of course, what made the extensions seem like an endless morass is that all the problem extensions we had just listed?six digits, seven,

thirty-three, three factors instead of two?were

special cases, just like the original problem. We had not really stopped to ask what they were special cases of.

The author has highlighted some insights into

multiplication that she gained while analyzing this discrete optimization problem. The generalization that comes directly from the author's analysis is this: Given n digits, how can we arrange them into m factors to produce the greatest (lowest) product? It seems perfectly fair to fix n or m for the analysis, but fixing both would make it too much of a special case again. But the author's focus on optimization in the context of elementary arithmetic reminded us of a problem in an old elementary textbook,

Math Workshop, by Robert Wirtz, Morton Botel, Max Beberman, and W. W. Sawyer (Chicago: Encyclopaedia Britannica Press, 1964): Given some whole number, like 12, find a way to parti tion it?say (6, 6) or (3, 4, 5) or (1, 1, 1, 9)?that gives a set of numbers with the maximum product. The products for the examples are 36, 60, and 9, respectively, so (3, 4, 5) seems the best set, but

perhaps there is one even better. However, finding the maximum product for partitions of 12 is not

interesting. What can we say in general about such a partitioning for n? What if we are not restricted to integers as we look for numbers whose sum is n? The ideas that arise from this investigation touch mathematics that students in the final years of high school are just beginning to approach.

REFERENCES Brown, Stephen L, and Marion I. Walter. Problem

Posing: Reflections and Applications. Hillsdale, NJ: Lawrence Erlbaum, 1993.

Fisher, Ann. Math Mind Stretchers. Grand Rapids, MI: Instructional Fair, 1997.

Hanna, Gila. "Proofs That Prove and Proofs That

Explain." In Proceedings of the International Group for the Psgchologg of Mathematics Education, Vol ume II, edited by G. Vergnaud, J. Rogalski, and M.

Artigue, pp. 45-51. Paris, 1989. ~

CHRISTY D. GRAYBEAL, [email protected], is ^^^H a former middle school mathematics teacher. ^^^^H

She is currently a doctoral candidate in curricu- ^^^^H

lum and instruction with an emphasis on mathe- ^^^^H matics education at the University of Maryland, ^^^^H College Park, MD 20742.

^^^H




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Mathematical Discourse || Whole Number Multiplication: There's More to It Than Might Be Expected!