assignment 26 qa

#$&*

course mth151

12/17/14 11

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

.

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:

the prime factorizations have 3*5 in common. 3*5=15 which is the greatest common divisor of 60 and 105

the prime factorization of 60 already has the 3*5 thats in the 105 prime factorization so we just need to add

the *7 to the 60 prime factorization

2*2*3*5*7=420

420 is the smallest number that contains both factorizations and is the least common multiple of 60 and 105

confidence rating #$&*:3

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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.

Alternative explanation:

As long as the number in the second question contains factors 3, 5 and 7 it will be a multiple of 105.

As long as it contains a factors 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 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 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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Self-critique (if necessary):ok

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Self-critique Rating:ok

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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

2*42

2*2*21

2*2*3*7

126

2*63

2*9*7

2*3*3*7

combine what is common to both prime factorizations

2*3*7=42 is the greatest common divisor

2*2*3*3*7=252 is the least common multiple

confidence rating #$&*:3

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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.

Alternative answer to the question of the LCM:

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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Self-critique (if necessary):ok

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Self-critique Rating:ok

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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*252

2*2*126

2*2*2*9*7

2*2*2*3*3*7

378

2*189

2*3*63

2*3*9*7

2*3*3*3*7

2*3*3*7=127 is the greatest common divisor

2*2*2*3*3*3*7=1512 is the least common multiple

confidence rating #$&*:3

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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:

220

2*110

2*2*55

2*2*5*11

726

2*363

2*3*121

2*3*11*11

2*11=22 greatest common divisor

2*2*3*5*11*11=7260 is the least common multiple

confidence rating #$&*:

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Self-critique Rating:ok

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