course Mth 151 ???????~~????Qv?j?assignment #024
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21:10:58 5.2.6 does 17 + 51 verify Goldbach for 68
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RESPONSE --> This is not true. 51 is not a prime, and both numbers must be prime.
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21:12:54 query 5.2.20 if 95 abundant or deficient?
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RESPONSE --> Ok to the explanation of number 6. I think I hit the next question button too many times. Answer for number 20: 95 1 + 5 + 19 = 25 25 < 95 deficient
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21:13:00 **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. **
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RESPONSE --> ok
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21:14:35 5.2.36 p prime and a, p rel prime then a^(p-1) - 1 div by p
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RESPONSE --> 3^(5-1) - 1= 80 80/5 = 16 2^(7-1) - 1 = 63 63/7 = 9
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21:14:56 ** 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. **
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RESPONSE --> ok
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21:16:46 query 5.2.42 does the nth perfect number have n digits?
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RESPONSE --> I don't think I understand what the ""nth"" perfect number means. I know that n stands for any number and I understand what n digits means, but I don't understand the first part.
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21:18:00 ** 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. **
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RESPONSE --> OK - I get it now. I thought it meant the next perfect number and I haven't a clue about that. Once I see the answer I see how easy that was.
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