course MTH 151
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21:55:04 `q001. There are three questions in this assignment. 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105. What do the prime factorizations of 60 and 105 having common? What is the prime factorization of the smallest number which contains within its prime factorization the prime factorizations of both 60 and 105?
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RESPONSE --> The prime factorizations of these numbere have 3 & 5 in common. I'm not exactly sure what this questions is asking. The prime factorization of the smallest number which is 60, is 2 x 3 x 5.
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21:57:14 The prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 have in common the product 3 * 5 = 15. This is the largest number that will divide evenly into both 60 and 105, and is called the greatest common divisor of 60 and 105. In order to contain to both of the prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 a number must contain in its prime factorizations the entire prime factorization 2 * 2 * 3 * 5, and in addition the 7 still necessary in order to contain 3 * 5 * 7. Thus the number must be 2 * 2 * 3 * 5 * 7 = 420. This number is a multiple of both 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 120, and is the smallest number which is a multiple of both. We therefore call 420 the Least Common Multiple of 60 and 105.
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RESPONSE --> Oh, ok. I think the wording of the question threw me off track. This is exactly what I have on my paper. I understand what you have shown here.
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22:06:51 `q002. What are the prime factorizations of 84 and 126, and how can they be used to find the greatest common divisor and the least common multiple of these two numbers?
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RESPONSE --> 84 = 2 x 42 = 2 x 2 x 21 = 2 x 2 x 3 x 7 126 = 2 x 63 = 2 x 7 x 9 = 2 x 7 x 3 x 3 = 2 x 3 x 3 x 7 LCM = 2^3 x 3^3 x 7^2 = 10,584 GCD = 126 / 84 = 1 Rem 42 84 / 42 = 2 Rem 0 GCD = 42
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22:10:05 The prime factorization of 84 is 2 * 2 * 3 * 7, and the prime factorization of 126 is 2 * 3 * 3 * 7. The greatest common divisor of these numbers is the number we build up from all the primes that are common to both of these prime factorizations. The two prime factorizations having common 2, 3 and 7, which give us the greatest common divisor 2 * 3 * 7 = 42. The least common multiple is made up of just those primes which are absolutely necessary to contain the two given numbers. This number would have to contain the first number 2 * 2 * 3 * 7, and would in addition need another 3 in order to contain 2 * 3 * 3 * 7. The least common multiple is therefore 2 * 2 * 3 * 3 * 7 = 252.
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RESPONSE --> OK, I came up with the GCD in a different manner, but I got it right. I keep trying to use all of the prime factorizations instead of what they have in common. My LCM was way off. I did not combine my prime factorizations correctly. I do understand what I have missed.
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22:19:05 `q003. Find the greatest common divisor and least common multiple of 504 and 378.
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RESPONSE --> The GCD is: 504 / 378 = 1 Rem 126 378 / 126 = 3 Rem ) GCD = 126 LCM = 504 = 2 x 2 x 2 x 3 x 3 x 7 378 = 2 x 3 x 3 x 3 x 7 = 2 x 3 x 7 LCM = 42
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22:21:55 We find that 504 = 2 * 2 * 2 * 3 * 3 * 7 and 378 = 2 * 3 * 3 * 3 * 7. The greatest common divisor can contain a single 2 since 378 has only a single 2 in its factorization, two 3's since both numbers contain at least two 3's, and a single 7. The greatest common divisor is therefore 2 * 3 * 3 * 7 = 126. The least common multiple must contain the first number, 2 * 2 * 2 * 3 * 3 * 7, and another 3 because of the third 3 in 378. The least common multiple is therefore 2 * 2 * 2 * 3 * 3 * 3 * 7 = 1512.
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RESPONSE --> Ok, I still got he GCD correct but I'm still missing the LCM. OH, I think I realize what I've been missing. I was missing the fact that you have to use the entire first number and then add anything else extra that is in the second number. I think I get this now.
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$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ S~hyCܦ}ƴQ߫ assignment #025 g}ၳQ} Liberal Arts Mathematics I 12-13-2005
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21:02:37 ** The prime factorizations are 180=2 ^2 * 3 ^ 2 * 5 and 300=2 ^2 * 3 ^1 * 5^2. They have in commin 2^2, 3 and 5, and no higher power of any of these factors. Since 2^2 * 3^1 * 5^1=60 the greatest common factor is 60. **
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RESPONSE --> Oops! I hit but button again. OK, this is exactly how I came to my conclusion and I have the same answer. I understand that to get the greatest common factor, you have to find the prime factorization of the numbers and then see what they have in common.
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21:16:56 query 5.3.24 Euclidean algorithm to find GCF(25,70) Show how you used the Euclidean algorithm to find the greatest common factor of the two numbers.
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RESPONSE --> To find the GCF, you would have to: 70 / 25 = 2 Rem 20 25 / 20 = 1 Rem 5 20 / 5 = Rem 4 The GCF is 5.
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21:17:15 ** To apply the Euclidean algorithm we divide the larger number by the smaller, obtaining a remainder. We then divide the remainder by the divisor and repeat this process until we get 0 remainder. The greatest common divisor is the last divisor. In this case 70 divided by 25 gives us remainder 20. Then we divide the previous divisor 25 by the remainder 20, obtaining remainder 5. Then we divide the previous divisor, which is now 20, by the remainder 5. The remainder of this division is 0. So the last divisor, which is 5, is the greatest common factor. **
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RESPONSE --> Ok, I got this one.
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21:27:59 query 5.3.36 LCM of 24, 36, 48 How did you use the prime factors of the given numbers to find their greatest common factor?
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RESPONSE --> The prime factorization for: 24 = 2^3 x 3 36 = 2^2 x 3^2 48 = 2^4 x 3 Which gives you 2 ^9 x 3 ^4=13,824 which is the greatest common factor.
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21:29:44 ** The prime factorizations are 24 = 2*2*2*3, 36 = 2*2*3*3, 48 = 2*2*2*2*3. The smallest number that includes all these factors has four 2's and two 3's. 2*2*2*2 * 3*3 = 144. So 144 is the GCF. **
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RESPONSE --> I think my mistake was that I added all of the prime factorizations instead of the largest in each group. I understand my mistake and the answer given here.
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21:36:06 query 5.3.48 GCF of 48, 315, 450 Show how you used the Euclidean algorithm to find the greatest common factor of the three given numbers.
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RESPONSE --> 48 315 450 -------------------- 3 = 16 105 150 16 x 105 = 1680 which is the GCF
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21:41:52 ** Applying the Euclidean Algorithm to 315 and 48: 315 divided by 48 gives us remainder 27. 48 divided by 27 gives us remainder 21. 27 divided by 21 gives us remainder 3. 6 divided by 3 gives us remainder 0. The last divisor is 3, which is therefore the GCF of 315 and 48. The GCF of the three numbers is therefore the GCF of 450 and 3, which is found by first dividing 450 by 3, which gives us remainder 0. So the last divisor is 3, which is therefore the GCF of the three numbers. **
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RESPONSE --> Ok, I was looking in the book at the chart for finding the GCF (Dividing by Prime Factors Method). I do understand here how to find the GCF for 3 numbers instead of 2. You have to find the GCF for the first 2 numbers and then find it for that number and the last number.
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21:43:13 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Well, I made a few errors. Most of which were silly mistakes but the last problem, the one on finding the GCF for 3 numbers threw me off for a second. I'm not sure why I deceded to go by the chart in the book that I told you about earlier. But I do understand how to do that now.
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