QA 24

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course Mth151

3/27 8:45

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.

024. More number theory

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Question: `q001. There are seven questions in this assignment.

Pick any even number--say, 28. It is believed that whatever even number you pick, as long as it is at least 6, you can express it as the sum of two odd prime numbers. For example, 28 = 11 + 17.

Express 28 as a some of two prime factors in a different way.

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Your solution: 5+23=28

confidence rating #$&*: 3

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Given Solution:

28 can be expressed as 5 + 23, both of which are prime.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q002. The assertion that any even number greater than 4 can be expressed as a sum of two primes is called Goldbach's conjecture. Verify Goldbach's conjecture for the numbers 42 and 76.

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Your solution: 42 can be expressed 31+11 both of which are primes.

76 can be expressed 73+3 both of which are primes.

confidence rating #$&*: 3

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Given Solution:

42 = 23 + 19, or 13 + 29, or 11 + 31, or 5 + 37.

76 = 73 + 3, 71 + 5, 59 + 17, 53 + 23, or 29 + 47.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q003. The proper factors of a number are the factors of that number of which are less than the number itself. For example proper factors of 12 are 1, 2, 3, 4 and 6.

List the proper factors of 18 and determine whether the sum of those proper factors is greater than, less than, or equal to 18 itself.

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Your solution: 1, 2, 3, 6, and 9. The sum of these numbers is 21 which is larger than 18.

confidence rating #$&*: 3

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Given Solution:

The proper factors of 18 are easily found to be 1, 2, 3, 6 and 9. When these factors are added we obtain 1 + 2 + 3 + 6 + 9 = 21. This result is greater than the original number 18.

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: `q004. A number is set to be abundant if the sum of its proper factors is greater than the number. If the sum of the proper factors is less than the number than the number is said to be deficient. If the number is equal to the sum of its proper factors, the number is said to be perfect.

Determine whether each of the following is abundant, deficient or perfect: 12; 26; 16; 6.

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Your solution: The factors of 12 are 1, 2, 3, 4, and 6 and their sum is 16. So 12 is abundant.

The factors of 26 are 1, 2, and 13 and their sum is 16. So 26 is deficient.

The factors of 16 are 1, 2, 4, and 8 and their sum is 15. So 16 is deficient.

The factors of 6 are 1, 2, and 3 and their sum is 6. So 6 is perfect.

confidence rating #$&*: 3

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Given Solution:

The proper factors of 12 are 1, 2, 3, 4 and 6. These proper factors add up to 16, which is greater than 12. Therefore 12 is said to be abundant.

The proper factors of 26 are 1, 2, and 13. These proper factors add up to 16, which is less than 26. Therefore 26 is said to be deficient.

The proper factors of 16 are 1, 2, 4 and 8. These proper factors add up to 15, which is less than 16. Therefore 16 is said to be deficient.

The proper factors of 6 are 1, 2, and 3. These proper factors add up to 6, which is equal to the original 6. Therefore 6 is said to be perfect.

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: `q005. There is a perfect number between 20 and 30. Find it.

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Your solution:

21: 1, 3, and 7- deficient

22: 1, 2, and 11 - deficient

23: 1 - deficient

24: 1, 2, 3, 4, 6, 8, and 12 - abundant

25: 1 and 5 - deficient

26: 1, 2, and 13 - deficient

27: 1, 3, and 9 - deficient

28: 1, 2, 4, 7, and 14. - perfect!!

29: 1 - deficient

28 is the number

confidence rating #$&*: 3

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Given Solution:

The numbers 23 and 29 are prime, and no prime number can be perfect (think about this for a minute and be sure you understand why).

20 has proper factors 1, 2, 4, 5 and 10, which add up to 22, so 20 is abundant and not perfect.

21 has proper factors 1, 3 and 7, which add up to 11, which make 21 deficient.

22 has proper factors 1, 2 and 11, which add up to 14, so 22 is deficient.

24 has proper factors 1, 2, 3, 4, 6, 8 and 12, which add up to 35, so 24 is abundant.

25 has proper factors 1 and 5, and is clearly deficient.

26 was seen earlier to be deficient.

27 has proper factors 1, 3 and 9, and is clearly deficient.

28 has proper factors 1, 2, 4, 7 and 14. These add up to 28. So 28 is a the perfect number we are looking for.

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Self-critique (if necessary):

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Self-critique Rating: 3

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Question: `q006. Why can't a prime number be perfect?

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Your solution: Because the only factor that it has besides itself is 1. 1 cannot add up to the number itself so it is therefore deficient.

confidence rating #$&*: 3

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Given Solution:

A prime number has only two factors, itself and 1. It therefore has only one proper factor, which is 1. Since every prime number is greater than 1,no prime number can be perfect.

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Self-critique (if necessary): I like your wording better. Makes more sense.

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Self-critique Rating: 3

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Question: `q007. 2^2 - 1 = 3, which is prime. 2^3 - 1 = 7, which is prime. 2^5 - 1 = 31, which is prime. Is it true that for any n > 1, 2^n - 1 is prime?

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Your solution: NO. There has been no formula developed that will work for any prime number.

confidence rating #$&*: 3

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Given Solution:

You might at first think that the above examples establish a pattern, but if you check n = 4 you find that 2^n - 1 = 2^4 - 1 = 16 - 1 = 15, which is divisible by 3 and 5 and is not prime.

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Self-critique (if necessary): Yes, I suppose I should’ve used an example.

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Self-critique Rating: 2

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Question: `q007. Check for p = 3, then p = 5, then p = 7 to see whether the formula 2^p - 1 seems to give primes.

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Your solution: 2^3-1 = 7 Prime

2^5-1 = 31 Prime

2^7-1 = 128 It is even therefore not prime. The only even prime number is 2 itself.

confidence rating #$&*: 3

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Given Solution:

2^3 - 1 = 8 - 1 = 7.

2^5 - 1 = 32 - 1 = 31.

2^7 - 1 = 128 - 1 = 127.

All these results are prime. However this doesn't prove that the formula always works. Your book will address this question.

Oh, Shoot!!! I forgot to subtract the 1!!

Self critique rating: 1

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q007. Check for p = 3, then p = 5, then p = 7 to see whether the formula 2^p - 1 seems to give primes.

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Your solution: 2^3-1 = 7 Prime

2^5-1 = 31 Prime

2^7-1 = 128 It is even therefore not prime. The only even prime number is 2 itself.

confidence rating #$&*: 3

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Given Solution:

2^3 - 1 = 8 - 1 = 7.

2^5 - 1 = 32 - 1 = 31.

2^7 - 1 = 128 - 1 = 127.

All these results are prime. However this doesn't prove that the formula always works. Your book will address this question.

Oh, Shoot!!! I forgot to subtract the 1!!

Self critique rating: 1

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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Good job, despite a couple of oversights.

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