finding the least common multiple
TRANSCRIPT
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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