problem solving using gcd's and lcm's of rational numbers

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Problem Solving Using GCD’s and LCM’s of Rational Numbers Ronald Edwards Westfield State College Westfield, Massachusetts 01085 Students in arithmetic classes become familiar with one or more al- gorithms for finding greatest common divisors (GCD’s) and least com- mon multiples (LCM’s) of given sets of natural numbers. Today these concepts are not extended to sets of rational numbers, but they can be and, indeed were in arithmetic texts of the nineteenth century. Practical computations in the nineteenth century often involved exten- sive work with fractions. Ownerships were divided and subdivided into quarters, eights, sixteenths, etc. . . Problems in inheritance, business transactions, measurement and currency involved complex work with fractional parts. The following four problems are examples (from The National Arithmetic, Benjamin Greenleaf, Robert S. Davis and Co., 1864) involving fractions: Problem 1. What is the smallest sum of money with which I could purchase a number of sheep at $2 1/4 each, a number of calves at $4 1/2 each and a number of yearlings at $9 3/8 each? And how many of each could I purchase with this money? Problem 2. I have three fields; the first contains 73 7/11 acres, the second 88 4/11 acres, the third 139 10/11 acres. Required, the largest sized house lots of the same extent into which the three fields can be di- vided, and also the number of lots. Problem 3. There is a certain island 80 miles in circumference. A, B and C agree to travel round it. A can walk 31/2 miles in an hour, B 4 2/3 miles, and C 5 1/4 miles. They start from the same point and continue their traveling 8 hours a day, until they shall all meet at the point from which they started. In how many days will they meet, and how far will each have traveled? Problem 4. My field has four sides. The first side is 31 rods 13 3/10 feet in length; the second, 41 rods 1 9/10 feet; the third, 38 rods 0 1/5 feet; the fourth, 45 rods 12 7/10 feet. I wish to enclose this field with a rail-fence four rails high, using rails of equal length. Required the length of the longest rails that can be used, allowing that the rails lap by each other 7/10 of a foot; also the number of rails it will take to fence it. First, the standard computations done in early arithmetic texts involv- ing GCD’s and LCM’s of fractions will be considered. These algorithms will be stated as theorems and proved. Finally, the techniques discussed will be applied in solving the four problems stated previously. 670

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Page 1: Problem Solving Using GCD's and LCM's of Rational Numbers

Problem Solving Using GCD’s andLCM’s of Rational Numbers

Ronald Edwards

Westfield State CollegeWestfield, Massachusetts 01085

Students in arithmetic classes become familiar with one or more al-gorithms for finding greatest common divisors (GCD’s) and least com-mon multiples (LCM’s) of given sets of natural numbers. Today theseconcepts are not extended to sets of rational numbers, but they can beand, indeed were in arithmetic texts of the nineteenth century.

Practical computations in the nineteenth century often involved exten-sive work with fractions. Ownerships were divided and subdivided intoquarters, eights, sixteenths, etc. . . Problems in inheritance, businesstransactions, measurement and currency involved complex work withfractional parts. The following four problems are examples (from TheNational Arithmetic, Benjamin Greenleaf, Robert S. Davis and Co.,1864) involving fractions:Problem 1. What is the smallest sum of money with which I could

purchase a number of sheep at $2 1/4 each, a number of calves at $4 1/2

each and a number of yearlings at $9 3/8 each? And how many of eachcould I purchase with this money?Problem 2. I have three fields; the first contains 73 7/11 acres, the

second 88 4/11 acres, the third 139 10/11 acres. Required, the largestsized house lots of the same extent into which the three fields can be di-vided, and also the number of lots.Problem 3. There is a certain island 80 miles in circumference. A, B

and C agree to travel round it. A can walk 31/2 miles in an hour, B 4 2/3

miles, and C 5 1/4 miles. They start from the same point and continuetheir traveling 8 hours a day, until they shall all meet at the point fromwhich they started. In how many days will they meet, and how far willeach have traveled?Problem 4. My field has four sides. The first side is 31 rods 13 3/10

feet in length; the second, 41 rods 1 9/10 feet; the third, 38 rods 0 1/5

feet; the fourth, 45 rods 12 7/10 feet. I wish to enclose this field with arail-fence four rails high, using rails of equal length. Required the lengthof the longest rails that can be used, allowing that the rails lap by eachother 7/10 of a foot; also the number of rails it will take to fence it.

First, the standard computations done in early arithmetic texts involv-ing GCD’s and LCM’s of fractions will be considered. These algorithmswill be stated as theorems and proved. Finally, the techniques discussedwill be applied in solving the four problems stated previously.

670

Page 2: Problem Solving Using GCD's and LCM's of Rational Numbers

Problem Solving 671

The Greatest Common Divisor

The greatest common divisor of two or more fractions is the greatestnumber that will divide each of them, and give a whole number for thequotient. In The National Arithmetic (pg. 183) Greenleaf states the fol-lowing rule:

Reduce the fractions, if necessary, to their least common divisor of the numerators,written over the least common denominator, will give the greatest common divisor re-quired.

Thus, to find the GCD of 4/15, 2 2/9 and 5 1/3 we may proceed as fol-lows: The LCM (15, 9, 3) = 45 and4/15 = 12/45,22/9 = 20/9 = 100/45, and5 1/3 = 16/3 = 240/45The GCD (12, 100, 240) = 4; Therefore, GCD (4/15, 2 2/9, 5 1/3) =4/45. Using the original fractions, we note: 4/15 - 4/45 =3,22/9-4/45 = 25, and 5 1/3 - 4/45 = 60.

The Least Common Multiple

The least common multiple (LCM) of two or more fractions is the leastnumber that can be divided by each of them, and give a whole numberfor the quotient. Greenleaf (pg. 184) states the following rule:

Reduce the fractions, if necessary, to their lowest terms. Then find the least commonmultiple of the numerators, which, written over the greatest common divisor of thedenominators, will give the least common multiple required.

Thus, to find the LCM of 4/5, 8/9 and 6/7 we may proceed as follows:The LCM (4,8,6) =24 andthe GCD (5,9,7) = 1. Therefore,LCM (4/5, 8/9, 6/7) = 24/1. Using the original fractions, we note: 24 -

4/5 = 30,24 - 8/9 = 27, and 24 - 6/7 = 28.

Three Theorems

Greenleaf’s rules are restated in theorems 1 and 2 below. Theorem 3 isan extension of the theorem: For natural numbers a and b, GCD (a,b) xLCM (a,b) = a x b. For natural numbers a,b,c,d we have:

Theorem 1. GCD (a/b, c/d) =GCD (^^LCM (b,d,)

Theorem 2. LCM (a/b, c/d) =LCM(^GCD (b,d)

Theorehi 3. GCD (a/b, c/d) x LCM (a/b, c/d) = ac/bd.The proof of theorem 1 follows:

Page 3: Problem Solving Using GCD's and LCM's of Rational Numbers

672 School Science and Mathematics

It is sufficient to show GCD (a/b, c/d) =GCD (a>c>)LCM (b,d)

where GCD (a,b) = GCD (c,d) = 1. Let x =GCD(a>c)

. We mustshow LCM(b,d)

(1) a/b - x and c/d - x are integers and (2) if a/b - r/s and c/d - r/sare integers and GCD (r,s) = 1, then x - r/s is an integer. Now, a/b - x

LCM (b,d)= a/b x

GCD (a,c)

Since GCD (a,c) divides a and b divides LCM (b,d), a/b - x is an inte-ger. Similarly, c/d - x is an integer. We have a/b - r/s = as/br is an in-teger and c/d - r/s = cs/dr is an integer implies r| as and r|cs. Since(r,s) = 1, we conclude r|a and r[c. Therefore, r|GCD(a,c). Also, b[asandd|cs implies b|s and d|s since GCD(a,b) = GCD(c.d) = 1.Therefore, LCM(b,d)|s. Thus,GCD (a,c) s .x �is an integer.LCM (b,d) r

Theorems 2 and 3 may be proved in a similar way. Also, theorems 1 and2 may be generalized for sets of more than two rationals.

Applications

Returning to problem #1, we wish to find the LCM (2 1/4, 4 1/2, 93/8). Using theorem 2 we findLCM (2 1/4, 4 1/2, 9 3/8) = 225/2 = 112 1/2. 112 1/2 -21/4= 50,112 1/2 -41/2= 25, and 112 1/2 -93/8= 12. Therefore, theamount required is $112 1/2 which could be used to buy 50 sheep, 25calves and 12 yearlings.

In problem #2, we must calculate the GCD (73 7/11, 88 4/11, 13910/11). Applying theorem I we find the solution: 41 lots of 7 4/11 acreseach.

In problem 3, we compute the time for one round trip of the island forA, B and C is 120/7, 320/21, and 160/7 hours, respectively. The LCM(120/7, 320/21, 160/7) = 960/7. This represents 960/7 hours of travel,or 5 5/7 days of travelling or 17 1/7 days travelling 8 hours per day. Inthis time A, B and C travel 480, 640 and 720 miles, respectively.Problem 4 is left as an exercise. The solution is: (note 1 rod = 161/2

feet) rail length, 13 1/2 feet; number of rails, 808.These algorithms and problems appearing in The National Arithmetic

suggest some interesting investigations into GCD’s, HCF’s, divisibilityand problem solving. These early texts provide a wealth of computa-tional techniques, practical problems, and insights into life and educa-tion in the 19th century.