FINDING THE LEAST COMMON MULTIPLEAuthor(s): Greg FreemanSource: The Mathematics Teacher, Vol. 76, No. 6 (September 1983), pp. 408-410Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27963578 .

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While students are completing the cards and studying the transparency, pass out manila folders on which the students write their names. Then give each student a copy of the course overview, classroom rules, and the seating arrangement (a duplicate of the transparency).

Carefully explain that this seating ar

rangement is probably different from any they have seen before, because their seats are not on the "blocks" of the seating ar

rangement but on the intersections of the lines. Explain that one of the important concepts of algebra is that of coordinate

graphing. Then tell the class about the kinds of geometric problems that led Ren? Descartes to the development of the rec

tangular coordinate system. Review the

concept of the number line and demonstrate on the chalkboard two number lines per pendicular to each other. Discuss with the class the methods by which a location on a Cartesian plane can be assigned a numeri cal value. Tell the class that the usual way to do this is through the ordered pair, in which the first value represents a move ment parallel to the x-axis (the horizontal number line) and the second value repre sents a movement parallel to the j>-axis (the vertical number line).

Pretend that the entire Cartesian plane has been transferred to the classroom floor. Demonstrate the location of (0, 0) and two or three other seats (preferably not as

signed to students) and ask the students to look at the upper-right corner of their index cards to find the directions to their per manent locations in the seating arrange ment. One or two students usually need

help, but try to let them help each other. Now that students are located in the

proper seats, ask them individually to rise, state and spell their names, and tell some

thing about themselves. This way, students will not be embarrassed by mispronunci ations or misspellings of their names. As

you write the names on the overhead tran

sparency, ask the students to write the names on their individual seating charts. See figure 3 for an example.

When the period is over, you've met the

Fig. 3. A completed seating chart

students, and they have met each other and

you. They have a copy of the course over

view, class rules, and seating charts, which

they are to keep in their folders. You can also use the folders as pieces of "mail" they pick up from a "mailbox" every day; the folders remaining in the box give an absen tee list. You can give students messages through their folders, and they keep their work in their folders. At the end of each class period, they return their folders to the box.

When this first class is over, you have an easily alphabetized class roster (the index cards), with all the information usu

ally needed right at your fingertips. You have an easy method for determining absen

tees, and you have a seating chart. Best of

all, you have a group of students who have internalized the concept of ordered pairs in the Cartesian rectangular coordinate

system.

Jean Avery Cocoa, FL 32922

FINDING THE LEAST COMMON MULTIPLE

Most students have difficulty finding the least common denominator (l.c.d.) or multi

ple when adding fractions. The prime facto rization method is sometimes too tedious, and the integral multiple method is often cumbersome (see fig. 1). Many students

multiply the denominators to find a

408 Mathematics Teacher

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(a) Prime factorization method

ifem. ?? 24o

(b) Integral multiple method

Fig. 1. Two common but tedious and cumbersome methods for finding the I.cm.

common denominator. This approach usu

ally results in a denominator that is much

larger and, therefore, more difficult to work with than the least common denominator. The following example illustrates a

method?one that is neither tedious nor

cumbersome?for finding the least common

multiple (Lem.) of two numbers.

Problem 1

Find the Lem. of 30 and 48.

30 15 5

8 48 24 Lem. = 2 ?3? (5 ?8) = 240.

The suggested method consists of (a) writ

ing the two numbers as a fraction, (b) sim

plifying the fraction, and (c) finding the

product of the factors that were removed

along with the numerator and the denomi nator of the reduced fraction.

A comparison of the suggested method with the two more commonly used methods reveals its simplicity. The prime factoriza tion method makes it necessary for the stu

dent to decide which factors to use in addi tion to writing each number as a product of

primes.

The suggested method works equally well in finding the Lem. of more than two numbers. When applying it to three num

bers, the factors can be removed from all three numbers at the same time or from any

pair of numbers.

Problem 2

Find the Lem. of 36,150, and 105

36 ?12 @12 ? 6 150 \ 50 10 / 5 105 \35 7

Lem.: 3 ? 5 ? 2 7) = 6300.

The prime factorization method would be

very involved for use in this example, and the integral multiple method would be com

pletely impractical. The suggested method of finding the Lem. has the following three benefits :

1. The greatest common factor (g.ef.) is the product of the factors that have been removed when finding the Lem. of two numbers. In problem 1, the g.ef. is

2-3 = 6.

The g.ef. of more than three numbers is found using this method by taking the prod uct of only those factors common to all the numbers.

2. The equivalent fractions with the Led. are easily found when adding frac tions. The student can look at the reduced fraction to determine what factors of the Led. are not factors of each denominator. See problem 3.

Problem 3

5 8 Find the sum of ? and ?.

36 63

September 1983

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Finding the l.c.d. yields

36_ 12_0 63 21 ? 3 ? 3 ?

(4 7) = 252.

A ' 32

63 [T]~252 35

5 ? ?=252

36 ? @ 67

252

3. The prime factorization of the num

bers can be readily found if the common

factors that are removed are restricted to

primes.

Greg Freeman

Independence High School

Independence, MS 38638

THE GOLF PROBLEM A favorite task to assign students in my course on computer programming is this :

Sal and Jessie are golfing partners. One

day they decide to play an eighteen-hole round of golf for the following wager :

1st hole $.01 2d hole $.02 3d hole $.04 4th hole $.08 5th hole $.16

18th hole

Write a BASIC program to determine the

amount wagered on the eighteenth hole.

Also, if Jessie wins all eighteen holes, what is the total amount he wins for the

day?

Most students will, without much diffi

culty, come up with a program to accom

plish the task. A sample BASIC program and its output are included in figure 1.

Students accustomed to searching for

patterns will make the following observa tions :

IO REM ?** THE GOLF PROBLEM ?*? 20 REM #? # *#** *?** # **?? ? **** **** ******** 30 REM ?* VARIABLES USED ?* 40 REM H- STORES HOLE NUMBER *** 50 REM ?? A- STORES AMOUNT WAGERED ?** GO REM *** AT EACH HOLE *** 70 REM * * S- STORES SUBTOTAL ** 80 REM ??fr******************************************* 90 PRINT "HOLE","AMOUNT "t"SUBTOTAL" 100 REM ****#*# * ###?##* ** ?## **#** *?**** *** 110 REM * INITIALIZE VARIABLES FOR HOLE #1 *** 120 LET H=l 130 LET A=.01 140 LET S=.01

150 PRINT "$"A *"*"S 160 REM * #*#** * ?# ****#?*?##?*** *#****? ****?****? 170 REM ** LOOP TO GENERATE AMOUNTS FOR REMAINING **? 180 REM ** HOLES *** 190 REM ** *****? *?#? * **#* ** *#*** # # ********* 200 FOR H= 2 TO IB 210 A=A*2 220 S=S+A

230 PRINT H,?'$"A,"S"S 240 NEXT H 250 REM ??*?#*#*# #* ***#*****?*** *#?**#**?* **?***** 300 END

HOLE AMOUNT SUBTOTAL 1 % .01 % .01 2 * .02 * .03 3 $ .04 * .07 4 S .08 $ .15 5 % .16 * .31 6 * .32 % .63 7 * .64 $ 1.27 8 * 1.2B % 2.55 9 * 2.56 * 5.11

10 t 5.12 * 10.23 11 * 10.24 S 20.47 12 * 20.48 * 40.95 13 * 40.96 % 81.91 14 % 81.92 % 163.83 15 % 163.84 t 327.67 16 * 327.68 % 655.35 17 % 655.36 * 1310.71 16 ? 1310.72 * 2621.43

Fig. 1

The amount wagered at each hole is a

power of 2, namely, 2H-1, where H is the number of the hole.

The subtotal at hole H is consistently shy of the amount wagered on the succeeding hole by a penny; that is, the subtotal at hole H is

2<H + n-i _ 1 = 2H - 1.

The amount wagered on hole 18 is 217

cents, or $1310.72, and the subtotal at hole 18 is (218

- 1) cents, or $2621.43.

These observations lead in a very natu

ral way to the development of the gener alization

2X~? =2"-I, = 1

or equivalently,

? 2* = 2 + 1 - 1. = 0

The output of the golf problem gives stu

dents empirical "evidence" of ? theorem

410 Mathematics Teacher

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