course Mth 151 Ry˕ገ卄Tпassignment #024
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21:09:25 5.2.6 does 17 + 51 verify Goldbach for 68
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RESPONSE --> no because 51 is not a prime number confidence assessment: 2
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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. **
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RESPONSE --> self critique assessment: 0
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21:14:20 query 5.2.20 if 95 abundant or deficient?
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RESPONSE --> 95 = 1,5,19 1+5+19 = 25 which is less than 95 so the answer is deficient confidence assessment: 2
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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. **
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RESPONSE --> self critique assessment: 0
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21:19:31 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 yes 80 is divisible by 5 2^(7-1) -1= 63 no 63 is not divisible by 5 confidence assessment: 2
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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. **
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RESPONSE --> self critique assessment: 0
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21:20:56 query 5.2.42 does the nth perfect number have n digits?
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RESPONSE --> I have no idea confidence assessment: 0
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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. **
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RESPONSE --> ok I see now self critique assessment: 2
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IzNiЍЕճᙦw assignment #025 025. `query 25 College Algebra 12-07-2008
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21:31:20 query 5.3.12 using prime factors find the greatest common factor of 180 and 300. What is the greatest common factor and how did you use prime factors to find it?
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RESPONSE --> 180=2^2*3^2*5 300=2^2*3*5^2 2^2*3*5 = 60 60 being GCF confidence assessment: 2
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21:31:31 ** The prime factorizations are 180=2 ^2 * 3 ^ 2 * 5 and 300=2 ^2 * 3 ^1 * 5^2. They have in commin 2^2, 3 and 5, and no higher power of any of these factors. Since 2^2 * 3^1 * 5^1=60 the greatest common factor is 60. **
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RESPONSE --> self critique assessment: 0
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21:48:51 query 5.3.24 Euclidean algorithm to find GCF(25,70) Show how you used the Euclidean algorithm to find the greatest common factor of the two numbers.
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RESPONSE --> Divide 70 by 25 and get the gcf of 5 confidence assessment: 2
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21:49:19 ** To apply the Euclidean algorithm we divide the larger number by the smaller, obtaining a remainder. We then divide the remainder by the divisor and repeat this process until we get 0 remainder. The greatest common divisor is the last divisor. In this case 70 divided by 25 gives us remainder 20. Then we divide the previous divisor 25 by the remainder 20, obtaining remainder 5. Then we divide the previous divisor, which is now 20, by the remainder 5. The remainder of this division is 0. So the last divisor, which is 5, is the greatest common factor. **
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RESPONSE --> self critique assessment: 0
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21:53:28 query 5.3.36 LCM of 24, 36, 48 How did you use the prime factors of the given numbers to find their greatest common factor?
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RESPONSE --> I divided 24, 36, and 48 by 2 and got 12,18,24 and so on and multiplied 2*2*3*2*3*4 = 288 my LCM is 288 confidence assessment: 2
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21:54:17 ** The prime factorizations are 24 = 2*2*2*3, 36 = 2*2*3*3, 48 = 2*2*2*2*3. The smallest number that includes all these factors has four 2's and two 3's. 2*2*2*2 * 3*3 = 144. So 144 is the GCF. **
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RESPONSE --> I thought we where suppose to find the LCM not the GCF, my bad self critique assessment: 2
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21:59:57 query 5.3.48 GCF of 48, 315, 450 Show how you used the Euclidean algorithm to find the greatest common factor of the three given numbers.
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RESPONSE --> divide 480 by 150 gives you 3 and divide 150 by 30 you get 5 with no remainder. so 3 would be the GCF confidence assessment: 2
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22:00:25 ** Applying the Euclidean Algorithm to 315 and 48: 315 divided by 48 gives us remainder 27. 48 divided by 27 gives us remainder 21. 27 divided by 21 gives us remainder 3. 6 divided by 3 gives us remainder 0. The last divisor is 3, which is therefore the GCF of 315 and 48. The GCF of the three numbers is therefore the GCF of 450 and 3, which is found by first dividing 450 by 3, which gives us remainder 0. So the last divisor is 3, which is therefore the GCF of the three numbers. **
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RESPONSE --> self critique assessment: 0
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22:00:43 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 0
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