course Mth 151 Ry˕ገ卄Tпassignment #024
......!!!!!!!!...................................
21:09:25 5.2.6 does 17 + 51 verify Goldbach for 68
......!!!!!!!!...................................
RESPONSE --> no because 51 is not a prime number confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:09:48 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
21:14:20 query 5.2.20 if 95 abundant or deficient?
......!!!!!!!!...................................
RESPONSE --> 95 = 1,5,19 1+5+19 = 25 which is less than 95 so the answer is deficient confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:14:40 **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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
21:19:31 5.2.36 p prime and a, p rel prime then a^(p-1) - 1 div by p
......!!!!!!!!...................................
RESPONSE --> 3^(5-1) -1= 80 yes 80 is divisible by 5 2^(7-1) -1= 63 no 63 is not divisible by 5 confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:19:52 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
21:20:56 query 5.2.42 does the nth perfect number have n digits?
......!!!!!!!!...................................
RESPONSE --> I have no idea confidence assessment: 0
.................................................
......!!!!!!!!...................................
21:21:39 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok I see now self critique assessment: 2
................................................."