Assignment 25

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course MTH 151

4/21/14, 11:02PM

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.

024. More number theory

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Question: `q001. There are 8 questions in this assignment.

Pick any even number--say, 28.

It is believed but not yet proven 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:

I picked the number 22. When 11, an odd prime number is added - 11 + 11, we get 22.

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:

23 + 19, both prime numbers, equals 42.

29 + 47, both primes, equals 76.

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, 9 are the factors of 18. 1 + 2 + 3 + 6 + 9 = 21, which is greater 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, 6, and they equal 16 when added. 16 is greater than 12, so it would be abundant. The factors of 26 are 1, 2, 13, and they equal 16, which is less than 26, so it would be deficient. The factors of 16 are 1, 2, 4, 8, which add up to 15, which is less than 16, making it deficient. The factors of 6 are 1, 2, 3, which add up to 6. 6 = 6, so it would be 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:

28, because its factors are 1, 2, 4, 7, and 14, which add up to 28. 28 = 28, so it would be considered perfect.

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):

Had to think about how a prime number could not be perfect, but I know now why.

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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 they are not divisible by anything other than 1.

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 had to think about this for quite a bit, and honestly may have overthought the question, but I do understand now.

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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, because it depends on the number being plugged in for n.

confidence rating #$&*: 3

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Given Solution:

You might at first think that the above examples establish a never-ending 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):

I was curious if this would work for just odd numbers, or for even and odd. Not sure if that’s really relevant or not, but it was just a thought.

@&
It's a good thought. However it turns out that 2^4 - 1 = 15, which is clearly not prime. Similarly, 2^6 - 1 = 63, which isn't prime.

You might begin to think that there's a pattern here, that with even numbers greater than 2 the calculation gives us non-primes; but then 28 - 1 = 127, which is prime.
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Self-critique Rating: 3

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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, where p stands for a prime number.

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Your solution:

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

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

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

All of these sums are prime numbers, but it depends on the number being plugged in for p as to what the outcome would be, and whether or not it would be prime.

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.

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Question: `q008. The number 2^p - 1, where p is the prime number 11, is itself a candidate for a prime number.

What is the value of 2^p - 1 for p = 11?

You aren't asked here to verify whether the number you have calculated is prime. However verify in the most efficient way you can, without the use of a calculator, whether 2, 3, 5 or 7 are divisors of your number.

Assuming that your candidate number is in fact prime, list the remaining numbers by which you would have to divide in order to verify this.

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Your solution:

2^11 = 2048; 2048 - 1 = 2047; 2047 / 2 = does not divide evenly because the number is odd, and would leave a decimal point. 2047 / 3 = 682? I don’t think the number would divide evenly either, I think it would have a decimal also. 2047 / 5 = 400? 2047 / 7 = 290?

I don’t think that 2047 is divisible by any of these numbers.

confidence rating #$&*:

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Self-critique Rating: 3

@&
Right.

As you will see when you study tests for divisibility:

2047 isn't divisible by 3 because the sume of its digits is 13, which isn't divisible by 3.

It isn't divisible by 5 because it doesn't end in 0 or 5.

It isn't divisible by 2 because it isn't even.

7 goes evenly into 2100, which differs from 2047 by 53. Since 53 isn't divisible by 7, neither is 2047.
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Assignment 25

&#Good responses. See my notes and let me know if you have questions. &#