#$&* course Mth 151 11/28/2013, 4:40 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** The Goldbach conjecture says that every even number greater than 2 can be expressed as the sum of two primes. 17 + 51 = 68 would verify the Goldbach conjecture except that 51 is not prime (51 = 3 * 17). So this sum does not verify the Goldbach conjecture. A sum that would satisfy the conjecture for 68 is 31 + 37 = 68, since 31 and 37 are both prime. COMMON ERROR AND INSTRUCTOR COMMENT: false 68 isn't a prime number Close, but 68 is the number being tested, which doesn't have to be prime (in fact since the conjecture addresses even numbers greater than two cannot be prime). The number being tested by the Goldback Conjecture is to be 'an even number greater than 2', which cannot be a prime number. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 5.2.20 if 95 abundant or deficient? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1,5,19, the sum of these is 25 deficient confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q5.2.36 p prime and a, p rel prime then a^(p-1) - 1 div by p YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3^(5-1)-1=3^4-1=81-1=80 80/5=16 verified 2^(7-1)-1=2^6-1=64-1=63 63/7=9 verified confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** This result is verified for both a=3, p=5 and a=2, p=7: If a = 3 and p = 5 then a and p have no common factors, so the conditions hold. We get a^((p-1))-1 = 3^(5-1) - 1 = 3^4 - 1 = 81 - 1 = 80. This number is to be divisible by p, which is 5. We get 80 / 5 = 16, so in this case a^(p-1)-1 is divisible by p. If a = 2 and p = 7 then a and p have no common factors, so the conditions hols. We get a^((p-1))-1 = 2^(7-1) - 1= 2^7 - 1 = 64 - 1 = 63. This number is to be divisible by p, which is 7. We get 63 / 7 = 9, so in this case a^(p-1)-1 is again divisible by p. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 5.2.42 does the nth perfect number have n digits? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes. The first 4 perfect numbers hold to that pattern, and that led to the conjecture that the nth perfect number contains n digits (page 233,235 in text),but after the fourth perfect number it does not hold true.(first 7 perfect numbers listed on pg 240 in text) 6, 1st perfect number, 1 digit 28, 2nd perfect numbe, 2 digits 496, 3rd perfect number, 3 digits 8128, 4th perfect number, 4 digits 33,550,336, 5th perfect number,8 digits pattern broken 8,589,869,056, 6th perfect number, 10 digits confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qquery 5.2.42 does the nth perfect number have n digits? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes. The first 4 perfect numbers hold to that pattern, and that led to the conjecture that the nth perfect number contains n digits (page 233,235 in text),but after the fourth perfect number it does not hold true.(first 7 perfect numbers listed on pg 240 in text) 6, 1st perfect number, 1 digit 28, 2nd perfect numbe, 2 digits 496, 3rd perfect number, 3 digits 8128, 4th perfect number, 4 digits 33,550,336, 5th perfect number,8 digits pattern broken 8,589,869,056, 6th perfect number, 10 digits confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!