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Prime Factorization and GCD-LCM Properties, Assignments of Trigonometry

University of Arizona (UA)Trigonometry

Solutions to homework problems related to finding prime factorizations, greatest common divisors (gcd), least common multiples (lcm), and verifying the equation gcd(a, b) × lcm(a, b) = a × b for different pairs of numbers.

Typology: Assignments

Pre 2010

Uploaded on 08/26/2009

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MATH 111 SOLUTIONS FOR HW 9
1. Write each of the following numbers as a product of primes:
•28 = 2 ×2×7 = 22·7
•48 = 2 ×2×2×2×3 = 24·3
•60 = 2 ×2×3×5 = 22·3·5
•91 = 7 ×13
•121 = 11 ×11 = 112
•1331 = 11 ×11 ×11 = 113
•10201 = 101 ×101 = 1012
•1030301 = 101 ×101 ×101 = 1013
2. Find the following greatest common divisors:
•GCD(28,60) = 22= 4
•GCD(28,91) = 7
•GCD(48,91) = 1
•GCD(121,1331) = 112= 121
3. Find the following least common multiples:
•LCM(28,60) = 22·3·5·7 = 420
•LCM(28,91) = 22·7·13 = 364
•LCM(48,91) = 24·3·7·13 = 4368
•LCM(121,1331) = 113= 1331
4. For each of the pairs (28,60), (28,91), (48,91), and (121,1331), verify the
equation GCD(a, b)×LCM(a, b) = a×b.
•GCD(28,60) ×LCM(28,60) = 4 ×420 = 1680 = 28 ×60
•GCD(28,91) ×LCM(28,91) = 7 ×364 = 2548 = 28 ×91
•GCD(48,91) ×LCM(48,91) = 1 ×4368 = 4368 = 48 ×91
•GCD(121,1331) ×LCM(121,1331) = 121 ×1331
5. Can you explain why the equation GCD(a, b)×LCM(a, b) = a×bis always
true?
Many of you wrote something like “Because every time we try it, it
works”; but this avoids the question. The thing we want to know is pre-
cisely, why does it work every time?
Others wrote that because
LCM(a, b) = a×b
GCD(a, b),
the equation GCD(a, b)×LCM(a, b) = a×bfollows. However, this is circular
reasoning: if you look at your notes from class, you will see that we con-
cluded LCM(a, b) = a×b
GCD(a,b)from the equation GCD(a, b)×LCM(a, b) =
a×bin the first place. So the question still remains: why is the equation
GCD(a, b)×LCM(a, b) = a×btrue?
pf2

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MATH 111 SOLUTIONS FOR HW 9

  1. Write each of the following numbers as a product of primes:
    • 28 = 2 × 2 × 7 = 2^2 · 7
    • 48 = 2 × 2 × 2 × 2 × 3 = 2^4 · 3
    • 60 = 2 × 2 × 3 × 5 = 2^2 · 3 · 5
    • 91 = 7 × 13
    • 121 = 11 × 11 = 11^2
    • 1331 = 11 × 11 × 11 = 11^3
    • 10201 = 101 × 101 = 101^2
    • 1030301 = 101 × 101 × 101 = 101^3
  2. Find the following greatest common divisors:
    • GCD(28, 60) = 2^2 = 4
    • GCD(28, 91) = 7
    • GCD(48, 91) = 1
    • GCD(121, 1331) = 11^2 = 121
  3. Find the following least common multiples:
    • LCM(28, 60) = 2^2 · 3 · 5 · 7 = 420
    • LCM(28, 91) = 2^2 · 7 · 13 = 364
    • LCM(48, 91) = 2^4 · 3 · 7 · 13 = 4368
    • LCM(121, 1331) = 11^3 = 1331
  4. For each of the pairs (28, 60), (28, 91), (48, 91), and (121, 1331), verify the equation GCD(a, b) × LCM(a, b) = a × b. - GCD(28, 60) × LCM(28, 60) = 4 × 420 = 1680 = 28 × 60 - GCD(28, 91) × LCM(28, 91) = 7 × 364 = 2548 = 28 × 91 - GCD(48, 91) × LCM(48, 91) = 1 × 4368 = 4368 = 48 × 91 - GCD(121, 1331) × LCM(121, 1331) = 121 × 1331
  5. Can you explain why the equation GCD(a, b) × LCM(a, b) = a × b is always true?

Many of you wrote something like “Because every time we try it, it works”; but this avoids the question. The thing we want to know is pre- cisely, why does it work every time? Others wrote that because

LCM(a, b) = a × b GCD(a, b)

the equation GCD(a, b)×LCM(a, b) = a×b follows. However, this is circular reasoning: if you look at your notes from class, you will see that we con- cluded LCM(a, b) = (^) GCD(a×ba,b) from the equation GCD(a, b) × LCM(a, b) = a × b in the first place. So the question still remains: why is the equation GCD(a, b) × LCM(a, b) = a × b true?

Let us look at a specific example first: GCD(28, 91) × LCM(28, 91) = 7 × (2^2 · 7 · 91), while 28 × 91 = (2^2 · 7) × (7 · 13). The same primes occur the same number of times each, in these products. For example, 2 occurs twice in the prime factorization of 28 and no times in 91; while 2 occurs no times in the GCD and twice in the LCM. The prime 7 occurs once each in 28 and 91, and it also occurs once each in the GCD and LCM. Let’s think about a slightly more abstract example. Suppose there’s a prime that occurs exactly five times in the prime factorization of a, and that this same prime occurs exactly three times in the prime factorization of b. (For example, a might be 96 = 2^5 · 3, and b might be 120 = 2^3 · 3 · 5. Then when a × b is written as a product of primes, the number of times that this prime occurs is 5 + 3 = 8. (For example, 96 · 120 = 2^8 · 32 · 5.) But when we calculate the GCD of a and b, we include this prime ex- actly three times (three being the smaller of three and five); for example, GCD(96, 120) = 2^3 · 3 = 24. Similarly, when we calculate the LCM, we in- clude this prime exactly five times; for example, LCM(96, 120) = 2^5 · 3 · 5 =

  1. So this prime also occurs 8 times in the product of the GCD and the LCM. In this manner, every prime occurs the same number of times in a · b as it does in GCD(a, b) · LCM(a, b), and so these two numbers are equal. Here’s a more mathematically complete solution, along these lines: sup- pose the prime p occurs k times in the prime factorization of a, and l times in the prime factorization of b. Then p occurs k + l times in the prime factorization of a · b. Also, the prime occurs in the prime factorization of GCD(a, b) a number of times equal to the minimum of k and l, and it occurs in the factorization of LCM(a, b) a number of times equal to the maximum of k and l. So p occurs in min(k, l) + max(k, l) times in the prime factor- ization of GCD(a, b) × LCM(a, b). But min(k, l) + max(k, l) = k + l, so p occurs in the prime factorizations of a · b and GCD(a, b) × LCM(a, b) the same number of times each. Therefore the prime factorizations of a · b and GCD(a, b) × LCM(a, b) are the same, and so these two numbers are equal.