course mth 151 assignment 22 was on the query file was not from 4.5 as stated in each question, it was from 4.6 so therefore because we were not assigned 4.6, my answers were wrong ??z?????o?m?w€??assignment #024024. `query 24 College Algebra 07-25-2008
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11:28:18 5.2.6 does 17 + 51 verify Goldbach for 68
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RESPONSE --> NO, BECAUSE 51 IS DIVISIBLE BY 17. 51=17*3 confidence assessment: 1
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11:29:02 ** 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 --> ok self critique assessment: 2
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11:34:15 query 5.2.20 if 95 abundant or deficient?
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RESPONSE --> 95=1,5,15,19 sum is 40, deficient confidence assessment: 1
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11:35:05 **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 --> i put 15 in but i knew it was deficient self critique assessment: 1
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11:44:05 5.2.36 p prime and a, p rel prime then a^(p-1) - 1 div by p
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RESPONSE --> p=5, a=3 for a^(p-1) -1 divided by p is: 3^(5-1) -1 3^4 (-1) 81-1 divided by 5 is 16 p=7, a=2 for a(p-1) -1 divided by p is 2^(7-1) 64-1 63 divided by 7 =9 confidence assessment: 0
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11:44:47 ** 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 self critique assessment: 2
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11:46:53 query 5.2.42 does the nth perfect number have n digits?
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RESPONSE --> no, not sure why confidence assessment: 0
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11:48:00 ** 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 self critique assessment: 0
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k???????????assignment #023 023. `query 23 College Algebra 07-25-2008
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18:59:57 5.1.18 List all the factors of 172.
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RESPONSE --> factors of 172 = 1,2,4,7,25,43,86,172 confidence assessment: 0
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19:01:02 ** If the number isn't too big we can simply start dividing by primes, beginning with the smallest: If we divide 172 by 2 we get 86, so 2 and 86 are factors. If we divide 172 by 3 we get 57 with a remainder so 3 isn't a factor. If we divide 172 by 4 we get 43, so 4 and 43 are factors. If we divide 172 by 5 we get 34 with a remainder so 5 isn't a factor. If we divide 172 by 6 we get 28 with remainder so 6 isn't a factor. If we divide 172 by 7 we get 24 with a remainder so 7 isn't a factor. If we divide 172 by 8 we get 21 with remainder so 8 isn't a factor. If we divide 172 by 9 we get 19 with a remainder so 9 isn't a factor. If we divide 172 by 10 we get 17 with a remainder so 10 isn't a factor. If we divide 172 by 11 we get 15 with a remainder so 11 isn't a factor. If we divide 172 by 12 we get 14 with a remainder so 12 isn't a factor. If we divide 172 by 13 we get 13 with a remainder so 13 isn't a factor. If we were to divide 172 by any number greater than 13 the result would be less than 13. We've already divided by every whole number less than 13 so we aren't going to find anything new by dividing by numbers greater than 13. Our factors are 2, 86, 4 and 43, as well as 1 and the number 172 itself. A method which is often quicker if the prime factorization contains a large number of factors is to list every prime factor, every product of two prime factors, every product of three prime factors, etc.: From the Prime Factorization 172 = 2 * 2 * 43 you find that the factors include: Each prime factor: 2 and 43 Each product of two prime factors: 2 * 2 = 4 and 2 * 43 = 86 The number itself and 1: 1 and 172. This method is quicker and more reliable than dividing by every possible number (what would you do with 5,668,725, for example?). **
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RESPONSE --> ok, add when using 7 trying to hurry self critique assessment: 1
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19:03:13 5.1.21 divisibility of 25025 by various factors. Explain how each divisibility test works for the number 25025.
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RESPONSE --> 25025 is divisible by 12,18,20,24 confidence assessment: 0
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19:04:32 ** 25025 is not divisible by 2 because it doesn't end in an even number. 25025 isn't divisible by 3 because the sum 2 + 5 + 0 + 2 + 5 = 14 of its digits is not divisible by 3. 25025 isn't divisible by 4 because its last two digits do not form a number divisible by 4. 25025 is divisible by 5 because its last digit is 5. 25025 isn't divisible by 6 because it isn't divisible by 2 and 3. 25025 isn't divisible by 8 because its last three digits do not form a number divisible by 8. 25025 isn't divisible by 9 because the sum 2 + 5 + 0 + 2 + 5 = 14 of its digits is not divisible by 9. 25025 isn't divisible by 12 because it isn't divisible by both 3 and 4.**
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RESPONSE --> did problem wrong self critique assessment: 0
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19:07:51 5.1.33 What is the prime factorization of 360 and how did you get it?
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RESPONSE --> 360= 2*180 2*2*90 2*2*3*30 2*2*3*3*10 2*2*3*3*2*5 confidence assessment: 2
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19:08:08 ** We can follow the simplest method, dividing successively by the smallest possible prime: 360 / 2 = 180, so 2 is a prime factor. 180 / 2 = 90, so 2 is again a factor. 90 / 2 = 45, so 2 is again a factor. 45 can't be divided by 2 so we note that 2 occurs 3 times as a factor and try division by 3: 45 / 3 = 15, so 3 is a factor. 15 / 3 = 5, so 3 is again a factor. 5 is itself prime. It follows that 360 = 2 * 2 * 2 * 3 * 3 * 5, as can be easily checked by multiplication. Thus the prime factorization is 360 = 2^3 * 3^2 * 5. **
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RESPONSE --> ok self critique assessment: 2
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19:12:29 5.1.60 number of divisors of 2^4*3^4*5^2
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RESPONSE --> 1,2,4,5,8,9,16,25,27,81 confidence assessment: 0
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19:13:35 ** The powers are 4, 4 and 2. The number of possible factors is therefore (4 + 1) * (4 + 1) * (2 + 1) = 5 * 5 * 3 = 75. **
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RESPONSE --> ok self critique assessment: 0
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19:15:27 query 5.1.80 is 2*3*...*13+1 prime?
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RESPONSE --> 30031 confidence assessment: 0
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19:15:46 ** To test for primeness you have to divide the number by every prime up to and including its square root. Having done so, you will either find that one of these primes does divide the number, or you will find that none does. Either way you will be able to answer the question. The number we need to test is 2 * 3 * 5 * 7 * 11 * 13 + 1 = 30031. Note that even though this is a pretty good-sized number it's not that big a task to divide by all primes up to the square root. The square root of 30031 is less than 174 so we only have to divide by primes less than 174, and there aren't all that many of them. Besides if one of the numbers 'works' we can stop. In fact 30,031 is not prime. Dividing by the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, all lead to fractional results. But on the next prime 59 we hit paydirt because 30,031 = 59 * 509 so 30031 isn't prime. **
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RESPONSE --> ok self critique assessment: 1
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z??{?????????? assignment #022 022. `query 22 College Algebra 07-25-2008
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22:37:24 4.5.9 {-1,0,1} group on multiplication?
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RESPONSE --> X -1 0 1 -1 -2 -1 0 0 -1 0 1 1 0 1 2 communtative and inverse confidence assessment: 0
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22:42:01 ** There are four criteria for the group: closure, identity, inverse property, and associativity. The lack of any one of these properties means that the set and operation do not form a group. The set is closed on multiplication. The identity is the element that when multiplied by other elements does not change them. The identity for this operation is 1, since 1 * -1 = -1, 1 * 0 = 0 and 1 * 1 = 1. Inverses are pairs of elements that give you 1 when you multiply them. For example -1 * -1 = 1 so -1 is its own inverse. 1 * 1 = 1 so 1 is also its own inverse. However, 0 does not have an inverse because there is nothing you can multiply by 0 to get 1. Since there is an element without an inverse this is not a group. **
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RESPONSE --> ok self critique assessment: 0
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22:46:27 4.5.25 verify (NT)R = N(TR)
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RESPONSE --> this is multiplication with associate property n=2 t=3 r=4 (2*3)*4= 2*(3*4) (6)4 = 2(12) 24=24 confidence assessment: 1
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22:47:21 ** From the table (NT)R= V R = M and N(TR)= N P = M This verifies the identity. **
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RESPONSE --> what table? self critique assessment: 0
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22:48:58 query 4.5.33 inverse of T
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RESPONSE --> r confidence assessment: 0
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22:49:33 ** T is its own inverse because T T gives you the identity **
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RESPONSE --> ok self critique assessment: 0
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22:51:50 4.5.42. Explain what property is gained when the system of integers is extended to the system of rational numbers.
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RESPONSE --> don't know confidence assessment: 0
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22:51:56 ** The set of integers is a group on addition, with identity 0 and every number x having additive inverse -x. It is not a group on multiplication. It contains the identity 1 but does not contain inverses, except for 1 itself. This is because, for example, there is no integer you can multiply by 2 to get the identity 1. If we extend the integers to the rational numbers we do get the inverses. The inverse of 2 is 1/2 since x * 1/2 = 1, the identity. In general the multiplicative inverse of x is 1 / x. However we still don't have a group on multiplication since 0 still doesn't have an inverse, 1 / 0 being undefined on the real numbers. **
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RESPONSE --> o self critique assessment: o
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