course Mth 151 E¤ð—™R“©¶Íã£]’à†è›Ù‹Ã´assignment #024
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15:03:40 `q001. There are seven questions in this assignment. Pick any even number--say, 28. It is believed that whatever even number you pick, as long as it is at least 6, you can express it as the sum of two odd prime numbers. For example, 28 = 11 + 17. Express 28 as a some of two prime factors in a different way.
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RESPONSE --> The sum of 13 and 15 is 28 confidence assessment: 2
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15:03:54 28 can be expressed as 5 + 23, both of which are prime.
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RESPONSE --> true self critique assessment: 2
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15:11:09 `q002. The assertion that any even number greater than 4 can be expressed as a sum of two primes is called Goldbach's conjecture. Verify Goldbach's conjecture for the numbers 42 and 76.
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RESPONSE --> 42 = 19+23 76 = 17+59 confidence assessment: 2
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15:11:35 42 = 23 + 19, or 13 + 29, or 11 + 31, or 5 + 37. 76 = 73 + 3, 71 + 5, 59 + 17, 53 + 23, or 29 + 47.
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RESPONSE --> OK self critique assessment: 2
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15:23:59 `q003. The proper factors of a number are the factors of that number of which are less than the number itself. For example proper factors of 12 are 1, 2, 3, 4 and 6. List the proper factors of 18 and determine whether the sum of those proper factors is greater than, less than, or equal to 18 itself.
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RESPONSE --> the prime factors of 18 is 1,2,3,6,9 and this adds up to 21 which is greater than 18 itself confidence assessment: 2
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15:24:18 The proper factors of 18 are easily found to be 1, 2, 3, 6 and 9. When these factors are added we obtain 1 + 2 + 3 + 6 + 9 = 21. This result is greater than the original number 18.
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RESPONSE --> that's what i got self critique assessment: 2
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15:30:27 `q004. A number is set to be abundant if the sum of its proper factors is greater than the number. If the sum of the proper factors is less than the number than the number is said to be deficient. If the number is equal to the sum of its proper factors, the number is said to be perfect. Determine whether each of the following is abundant, deficient or perfect: 12; 26; 16; 6.
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RESPONSE --> 12= 1,2,3,4,6 which = 16 making it abundant 26= 1,2,13 which =16 making it dificient 16 = 1,2,4,8 = 15 making it dificient 6 = 1,2,3 = 6 making it perfect confidence assessment: 2
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15:30:57 The proper factors of 12 are 1, 2, 3, 4 and 6. These proper factors add up to 16, which is greater than 12. Therefore 12 is said to be abundant. The proper factors of 26 are 1, 2, and 13. These proper factors add up to 16, which is less than 26. Therefore 26 is said to be deficient. The proper factors of 16 are 1, 2, 4 and 8. These proper factors add up to 15, which is less than 16. Therefore 16 is said to be deficient. The proper factors of 6 are 1, 2, and 3. These proper factors add up to 6, which is equal to the original 6. Therefore 6 is said to be perfect.
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RESPONSE --> self critique assessment: 0
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15:39:06 `q005. There is a perfect number between 20 and 30. Find it.
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RESPONSE --> The perfect number between 20 and 30 is 28 the proper no's of 28 is 1,2,4,7,14 which adds up to 28 confidence assessment: 2
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15:39:38 The numbers 23 and 29 are prime, and no prime number can be perfect (think about this for a minute and be sure you understand why). 20 has proper factors 1, 2, 4, 5 and 10, which add up to 22, so 20 is abundant and not perfect. 21 has proper factors 1, 3 and 7, which add up to 11, which make 21 deficient. 22 has proper factors 1, 2 and 11, which add up to 14, so 22 is deficient. 24 has proper factors 1, 2, 3, 4, 6, 8 and 12, which add up to 35, so 24 is abundant. 25 has proper factors 1 and 5, and is clearly deficient. 26 was seen earlier to be deficient. 27 has proper factors 1, 3 and 9, and is clearly deficient. 28 has proper factors 1, 2, 4, 7 and 14. These add up to 28. So 28 is a the perfect number we are looking for.
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RESPONSE --> self critique assessment: 0
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15:40:13 `q006. Why can't a prime number be perfect?
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RESPONSE --> because you don't include the number itself to proper factors confidence assessment: 2
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15:41:32 A prime number has only two factors, itself and 1. It therefore has only one proper factor, which is 1. Since every prime number is greater than 1,no prime number can be perfect.
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RESPONSE --> self critique assessment: 0
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15:43:51 `q007. 2^2 - 1 = 3, which is prime. 2^3 - 1 = 7, which is prime. 2^5 - 1 = 31, which is prime. Is it true that for any n > 1, 2^n - 1 is prime?
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RESPONSE --> this is not true if I took the number 4 and placed it for n you would get 15 and 15 is not a prime number. confidence assessment: 2
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15:44:10 You might at first think that the above examples establish a pattern, but if you check n = 4 you find that 2^n - 1 = 2^4 - 1 = 16 - 1 = 15, which is divisible by 3 and 5 and is not prime.
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RESPONSE --> self critique assessment: 0
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15:48:33 `q007. Check for p = 3, then p = 5, then p = 7 to see whether the formula 2^p - 1 seems to give primes.
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RESPONSE --> 3 gives you a prime, 5 gives you a prime and 7 gives you a prime confidence assessment: 2
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15:48:48 2^3 - 1 = 8 - 1 = 7. 2^5 - 1 = 32 - 1 = 31. 2^7 - 1 = 128 - 1 = 127. All these results are prime. However this doesn't prove that the formula always works. Your book will address this question.
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RESPONSE --> self critique assessment: 0
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ÏÞ´È°…ð¤°Æõ‰rÏççÀšðàÛÝœ assignment #025 025. GCF, LCM Liberal Arts Mathematics I 12-07-2008
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15:56:29 `q001. There are three questions in this assignment. 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105. What do the prime factorizations of 60 and 105 having common? What is the prime factorization of the smallest number which contains within its prime factorization the prime factorizations of both 60 and 105?
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RESPONSE --> I can't figure out what you are asking here? confidence assessment: 0
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16:00:22 The prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 have in common the product 3 * 5 = 15. This is the largest number that will divide evenly into both 60 and 105, and is called the greatest common divisor of 60 and 105. In order to contain to both of the prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 a number must contain in its prime factorizations the entire prime factorization 2 * 2 * 3 * 5, and in addition the 7 still necessary in order to contain 3 * 5 * 7. Thus the number must be 2 * 2 * 3 * 5 * 7 = 420. This number is a multiple of both 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 120, and is the smallest number which is a multiple of both. We therefore call 420 the Least Common Multiple of 60 and 105.
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RESPONSE --> I really don't see how 420 can be the least common multiple but I see how you are getting it
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16:33:24 `q002. What are the prime factorizations of 84 and 126, and how can they be used to find the greatest common divisor and the least common multiple of these two numbers?
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RESPONSE --> ok pf of 84 = 2*2*3*7 pf of 126= 2*3*3*7 confidence assessment: 2
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16:35:28 The prime factorization of 84 is 2 * 2 * 3 * 7, and the prime factorization of 126 is 2 * 3 * 3 * 7. The greatest common divisor of these numbers is the number we build up from all the primes that are common to both of these prime factorizations. The two prime factorizations having common 2, 3 and 7, which give us the greatest common divisor 2 * 3 * 7 = 42. The least common multiple is made up of just those primes which are absolutely necessary to contain the two given numbers. This number would have to contain the first number 2 * 2 * 3 * 7, and would in addition need another 3 in order to contain 2 * 3 * 3 * 7. The least common multiple is therefore 2 * 2 * 3 * 3 * 7 = 252.
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RESPONSE --> ok I see how to get the gcd and the lcm now self critique assessment: 2
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16:40:52 `q003. Find the greatest common divisor and least common multiple of 504 and 378.
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RESPONSE --> the prime factorizations of 504 AND 378 is 504 = 2^3*3^2*7 378 = 2^1*3^3*7 gcm is 42 and LCM im not for sure yet confidence assessment: 2
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16:41:53 We find that 504 = 2 * 2 * 2 * 3 * 3 * 7 and 378 = 2 * 3 * 3 * 3 * 7. The greatest common divisor can contain a single 2 since 378 has only a single 2 in its factorization, two 3's since both numbers contain at least two 3's, and a single 7. The greatest common divisor is therefore 2 * 3 * 3 * 7 = 126. The least common multiple must contain the first number, 2 * 2 * 2 * 3 * 3 * 7, and another 3 because of the third 3 in 378. The least common multiple is therefore 2 * 2 * 2 * 3 * 3 * 3 * 7 = 1512.
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RESPONSE --> ok self critique assessment: 2
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