Teaching Division to Reluctant Learners

Teaching Division to Reluctant LearnersAuthor(s): Harold E. ReesSource: The Mathematics Teacher, Vol. 72, No. 8 (NOVEMBER 1979), pp. 592-595Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27961800 .

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Example 2

Compute the quotient 389 17, and ex

press the remainder as a fraction.

Procedure. The usual sequence of steps to perform this quotient is given in figure 2

(see steps 1 and 2); the quotient is

22.88235294.

In steps 3 through 6, the integer part of the

quotient is subtracted off, and the differ ence is multiplied by 17, resulting in 15.

(See step 6.) Conclusion. 389 -f- 17 = 22 4- 15/17.

Remark 3 (Rationale). The result of di

viding 389 by 17 is an integer quotient 22 and a remainder of the form r/17, where r

is to be determined. The first two steps in

figure 2 indicate that the result of the divi sion is

22.88235294.

Therefore,

so that

22 + ? = 22.88235294,

r=.88235294 X 17 = 15.

Note 1. Although the steps given in fig ures 1 and 2 would not give correct results if performed on a calculator that does not have algebraic logic and on certain other

calculators, it should be apparent that they can be appropriately modified to accom modate such calculators.

Note 2. A more serious problem is one in which the calculator being used truncates its results rather than rounds them. On such a calculator, for example, the compu tation 1/3x3 results in the product

.99999999,

or, if a ten-digit calculator, .9999999999.

Enter Press

(1) 389 (2) 17 (3) No entry (4) 22 (5) No entry (6) 17

Display 389. 22.88235294 22.88235294 .8823529412 .8823529412 15.

Fig. 2

However, users of such calculators will have already had to accommodate such re sults and will know how to interpret them if they arise in examples of the kind consid ered above.

Concluding remark. In addition to pro viding a quick and practical way of obtain

ing fractional results from a calculator not

designed to do so, the procedure above af fords the teacher yet another opportunity to emphasize the important relationship between decimals and fractions.

Joseph Ercolano Baruch College, CUNY New York, NY 10010

Teaching Division to Reluctant Learners

Most mathematics teachers are well aware of the fact that division is repeated subtraction. For ten years I used the calcu lator as an aid to teach division as repeated subtraction. There was always a minimal amount of improvement in my classes' skills when they had to do division by hand, but the improvement never lived up

to my expectations. The students could not see the relationship between the repeated subtraction we did on the calculator and the normal division algorithm we use. Fi

nally I realized that I was going to have to use a different approach.

My classes are mostly tenth-grade stu dents who fall into one or more of the fol

592 Mathematics Teacher



lowing categories: (1) They had failed their ninth-grade mathematics and need a credit in mathematics in order to graduate; (2) they want to take more mathematics, but are not ready to take first-year algebra; (3) they have some type of learning disability; and (4) they are uninterested in school in

general and mathematics in particular. Many of them do not know the multiplica tion tables. All of them have a great deal of trouble performing division by hand.

The first step in preparing students for a new division algorithm begins by using the calculator to perform some repeated sub traction problems. In order to find the an swer to 103 -s- 7, they subtract 7 from 103, then from 96, 89, and so on, until they have a difference less than 7 (fig. 1). Then they

103 -7 96 -7 89 -7 82 -7 75 -7 68 -7 61 -7 54 -7 47 -7 40 -7 33 -7 26 -7 19

-7 12

-7 5

Fig. 1. 103 - 7 = 14 R5.

count the number of times that 7 was sub tracted. This number is the quotient. The last positive integer, in this case 5, is the re

mainder; therefore, 103 -s- 7 = 14 R5. Since this is a long procedure, it is im

portant not to pick problems with large quotients. After a few examples the proce dure can be shortened.

Using the problem 3697 -s- 3, I have the students begin to subtract multiples of 10 times 3; that is, they subtract 30, 300, 3000, ... (fig. 2). Since 3000 is the largest of these that does not exceed the dividend, it is the

3697 -3000 1000

697 -300 100

397 -300 100

97 -30 10

67 -30 10

37 -30 10

7 -3 1

4 -3 1

1

1232 R1

Fig. 2. 3697-5-3 = 1232 Rl.

first number they subtract. They subtract 3000 until the difference is less than 3000, then they subtract 300 until the difference is less than 300, then 30 until the difference is less than 30. Finally, they subtract 3 until

they get less than 3. Since 3000 is 1000 times 3, it represents 3

subtracted 1000 times; 300 is 3 subtracted 100 times; and 30 is 3 subtracted 10 times.

To the right of their work, the students write the number of times that 3 was sub tracted. Then at the bottom they add the number of times that 3 was subtracted. So 3697 + 3 equals 1232 Rl.

November 1979 593



It takes plenty of patience and planning to teach this procedure; four or five ex

amples are needed with one-digit divisors before going to two- and three-digit divi sors. The students need to do plenty of

problems if they are to succeed. After two or three classes they are ready to begin the new division algorithm.

I usually start by telling the students that I realize that many of them have trouble with division by hand and that the new

method will be strange to them at first but

they can improve by using it. I also have to insist that everyone use the method.

I begin with a problem with a one-digit divisor, such as 1734 -s- 5. My dialogue usu

ally goes as follows. I have to decide the largest multiple of

10 times 5 that can be subtracted from 1734

(fig. 3). The number is 500. Subtract 500

1734 -500 1234

-500 734

-500 234

-50

184 -50

134 -50

84

-50

34

-5

29

_^ 24

_^5 19

_^5 14

_^5 9

-5

4

100

100

100

10

10

10

10

346 R4

until the difference is less than 500, then subtract 50 until the difference is less than 50. Finally, subtract 5 until the difference is

less than 5. After each subtraction, record on the right of my funny-looking division box the number of times that 5 was sub tracted. I pretend that I am not sure how

many times 500, 50, or 5 can be subtracted. And above all, I go slowly. Several prob lems are worked until the students under stand the process is the same as what they were doing on the calculator. Next I begin to make the operation more concise. Whenever possible in the next few ex

amples, I subtract multiples of 2, 20, 200, ... times the divisor. I tell the students that I am getting a little bolder. If I am not sure

that the divisor can be subtracted 20 or 10

times, then I always choose the smaller number of times. A key to having success with this algorithm is to pretend that you are not good at division. See figure 4 for 2721 h-7 equals 388 R5.

After working several problems, I go on

to subtracting multiples of 3, 30, ... times

7 I 2721 -1400

1321 -700

621 -140

481 -140

341 -140

201 -140

61 -14

47

-14

33

-14

19

-14

200

100

20

20

20

20

2

2

2

_2_ 388 R5

Fig. 3. 1734 + 5 - 346 R4. Fig. 4. 2721 +7 = 388 R5.

594 Mathematics Teacher



the divisor. Usually higher multiples are not used. It's important to stress that I am

trying to teach this algorithm to people that are not good at estimating how many times

they can subtract the divisor; therefore, I

pretend that I have the same problem. My better students do not have as much

trouble making estimations of how many times to subtract the divisor. Their prob lems will look like figure 5. They will take the same number of steps as they would in the division algorithm that is normally taught. The only difference is where they

will write the quotient. When using this method, instructors

must demonstrate several examples for the students. They must pretend that they have

13463 -9000

4463 -3600

863 -810

53 -45

8

1000

400

90

_5 1495 R8

Fig. 5. 13463 -s- 9 = 1495 R8.

the same problems with division that their students have. It has long been a con tention of mine that students learn by imi tation. The instructor must let students ob serve a process over and over again. Too often we expect the students to pick up what we are showing them with only one or two examples.

This method has helped my students in several areas. Those who have trouble esti

mating the number of times a divisor will divide into the dividend learn to use a smaller estimation they are sure will work. If the estimation is too small, then they can subtract again to get an answer. Those who do not know the multiplication tables learn to subtract the divisor 1, 10, 100, or 1000 times. It may take them longer to do the

problem, but they will succeed. The num ber of place value errors is reduced, espe cially in problems with multiple zeros in the quotient.

Harold E. Rees Great Bend Senior High School Great Bend, KS 67530

Editor's Note. Readers interested in another alter

native to teaching division might wish to review "A

Duplation Method of Long Division" (November

1978).

A Problem in Probability

Consider the following problem: What is the probability of rolling a 12 with five 4-sided dice?

This work was completed by John Tkaczyk when

he was a student of Jesse Zoffer at Marlboro High

School, Marlboro, New Jersey. It was written before

the author had studied the permutation formula.

The answer is a fraction with a numerator

representing the number of possible combi nations of the dice resulting in a sum of 12

and a denominator representing all pos sible combinations of five 4-sided dice.

The denominator of the fraction can be

computed fairly easily. We rely on the fun

November 1979 595



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Teaching Division to Reluctant Learners