#$&*
course MTH 151
4/21/14, 11:27pm
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
025. GCF, LCM
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Question: `q001. There are 4 questions in this assignment.
2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105.
What do the prime factorizations of 60 and 105 having common?
What is the prime factorization of the smallest number which contains within its prime factorization the prime factorizations of both 60 and 105?
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Your solution:
Both have the numbers 3 and 5. I’m not really so sure what the second part of the question is asking? I’m having trouble understanding what it is asking for, because the smallest number for 60 is 2, and for 105 is 3.
@&
As long as the number in the second question contains 3, 5 and 7 it will be a multiple of 105.
As long as it contains a pair of 2's, a 3 and a 5 it is a multiple of 60.
If it satisfies both of these conditions then it is a multiple of both numbers.
The smallest number that satisfies both of these conditions contain 3 and 5, and also a pair of 2's (required so it can be a multiple of the first numbers), and also a 7 (required so it can be a multiple of the second number).
Thus the smallest number which is a multiple of both of the given numbers is
2 * 2 * 3 * 5 * 7.
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confidence rating #$&*: 2
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Given Solution:
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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Self-critique (if necessary):
I’m still a little confused on the second part of the question, and how you would know to multiply all the numbers together.
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Self-critique Rating: 2
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Question: `q002. What are the prime factorizations of 84 and 126, and how can these two prime factorizations be used to find the greatest common divisor and the least common multiple of these two numbers?
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Your solution:
84’s factors are 2 * 2 * 3 * 7.
126 = 2 * 3 * 3 * 7
The greatest common divisor would be 42 for both numbers. The LCM would be 2 * 2 * 3 * 3 * 7 = 252.
confidence rating #$&*:
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Given Solution:
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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Self-critique (if necessary):
For LCM, do you always just multiply all the numbers together? I’m a little confused with this.
@&
You get a number if you multiply its factors. What we do when we find the LCM is find the necessary factors; having done so we multiply them together.
You find what factors the two given numbers have in common. Since the LCM must be a multiple of both numbers, it must include these factors. In this case those factors would be 2, 3 and 7. Any number which does not have these factors cannot be a multiple of both 84 and 126.
Then you find what additional factors are necessary for your LCM to be a multiple of the first number. To be a multiple of the first number, the LCM must contain (in addition to 2, 3 and 7) another 2.
Then you find what additional factors are necessary for your LCM to be a multiple of the second number. To be a multiple of the second number, the LCM must contain (in addition to 2, 3 and 7) another 3.
Thus the LCM must contain 2, 3, 7, another 2, and another 3 as factors.
These factors are required, and they are also sufficient, so 2 * 2 * 3 * 3 * 7 is the LCM.
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Self-critique Rating: 3
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Question: `q003. Find the greatest common divisor and least common multiple of 504 and 378.
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Your solution:
504 - 2 * 2 * 2 * 3 * 3 * 7
378 - 2 * 3 * 3 * 3 * 7
GCD = 126
LCM = 1512?
@&
Those numbers are correct, but you haven't indicated how you reasoned them out, so I can't tell if you're doing this in a way that will always work.
*@
confidence rating #$&*: 2
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Given Solution:
The prime factorizations are
504 = 2 * 2 * 2 * 3 * 3 * 7 and
378 = 2 * 3 * 3 * 3 * 7.
The greatest common divisor can contain just 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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Question: `q004. Find the greatest common factor and the least common multiple of 220 and 726.
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Your solution:
GCF = 22
LCD = 1
@&
You don't show your reasoning here.
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confidence rating #$&*:
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Self-critique Rating: 1
@&
My answer to Question `q002 should have read as follows:
We are looking for the smallest number that is a multiple of both of the given numbers.
To be a multiple of some number, another number include all the factors of that number.
As long as the LCM includes 3, 5 and 7 as factors, it will be a multiple of 105.
As long as it contains a pair of 2's, a 3 and a 5 it is a multiple of 60.
If it satisfies both of these conditions then it is a multiple of both numbers.
The two given numbers both have 3 and 5 as factors.
The smallest number that satisfies both of these conditions therefore contains as factors 3 and 5, and also a pair of 2's (required so it can be a multiple of the first numbers), and also an additional 3 and a 7 (required so it can be a multiple of the second number).
Thus the smallest number which is a multiple of both of the given numbers has a pair of 2's, a pair of 3's, a 5 and a 7 as factors. This is the LCM, which is equal to
2 * 2 * 3 * 3 * 7.
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