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.  `query 24

Question:  `q5.2.6 does 17 + 51 verify Goldbach for 68

Your solution: 

Confidence Assessment:

Given Solution: 

`a** The Goldbach conjecture says that every even number greater than 2 can be expressed as the sum of two primes.

17 + 51 = 68 would verify the Goldbach conjecture except that 51 is not prime (51 = 3 * 17).

So this sum does not verify the Goldbach conjecture.

A sum that would satisfy the conjecture for 68 is 31 + 37 = 68, since 31 and 37 are both prime.

COMMON ERROR AND INSTRUCTOR COMMENT:  false   68 isn't a prime number

Close, but 68 is the number being tested, which doesn't have to be prime (in fact since the conjecture addresses even numbers greater than two cannot be prime).  The number being tested by the Goldback Conjecture is to be 'an even number greater than 2', which cannot be a prime number. **

Self-critique (if necessary):

Self-critique Rating:

Question:  `qquery 5.2.20 if 95 abundant or deficient?

Your solution: 

Confidence Assessment:

Given Solution: 

`a**The proper factors of 95 are 1, 5 and 19.

These proper factors add up to 25.

Since the sum of the proper factors is less than 95, we say that 95 is deficient. **

Self-critique (if necessary):

Self-critique Rating:

Question:  `q5.2.36 p prime and a, p rel prime then a^(p-1) - 1 div by p

Your solution: 

Confidence Assessment:

Given Solution: 

`a**  This result is verified for both a=3, p=5 and a=2, p=7:

If a = 3 and p = 5 then a and p have no common factors, so the conditions hold.  We get a^((p-1))-1 = 3^(5-1) - 1 = 3^4 - 1 = 81 - 1 = 80.

This number is to be divisible by p, which is 5.  We get 80 / 5 = 16, so in this case a^(p-1)-1 is divisible by p.

If a = 2 and p = 7 then a and p have no common factors, so the conditions hols.  We get a^((p-1))-1 = 2^(7-1) - 1= 2^7 - 1 = 64 - 1 = 63.

This number is to be divisible by p, which is 7.  We get 63 / 7 = 9, so in this case a^(p-1)-1 is again divisible by p. **

Self-critique (if necessary):

Self-critique Rating:

Question:  `qquery 5.2.42 does the nth perfect number have n digits?

Your solution: 

Confidence Assessment:

Given Solution: 

`a** The answer is 'no'.  The first perfect number, 6, has one digit. 

The second perfect number, 28, has 2 digits.  So far so good. 

The third perfect number is 496.  Still OK. 

The fourth is  8128, so we're still in good shape. 

But the fifth perfect number is 33,550,336, which has 7 digits, so the pattern is broken. 

The pattern never gets re-established.  Note that the sixth perfect number has ten digits. **