course mth 151 b͌W鑩ٍHxw{Student Name:
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14:18:29 `q001. There are twelve questions in this assignment. The number 12 is evenly divisible by 1, 2, 3, 4, 6, and 12. We say that 1, 2, 3, 4, 6 and 12 are the divisors of 12. Each of these divisors can be multiplied by another to get 12. e.g., 2 * 6 = 12, 1 * 12 = 12, 3 * 4 = 12. List the numbers from 2 to 20 and list all the divisors of each.
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RESPONSE --> the numbers 2-20 and there divisers 2= 1, 2 3= 1,3 4= 1,2,4 5=1,5 6-1,2,3,6 7=1,7 8=1,2,4,8 9=1,3,9 10=1,2,5,10 11=1,11 12=1,2,3,4,6,12 13=1,13 14=1,2,7,14 15=1,3,5,15 16=1,2,3,4,16 17=1,17 18=1,2,3,6,9,18 19=1,19 20=1,2,4,5,10,20
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14:20:22 `q002. Some of the numbers you listed have exactly two divisors. Which are these?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. numbers that have only 2 divisors are called prime numbers. prime numbers 2-20 are 2,3,5,7,11,13,17 and 19.
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14:21:35 `q003. These numbers with exactly two divisors are called prime numbers. List the prime numbers between 21 and 40.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Prime numbers between 21 and 40 are 23,29,31, and 37.
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14:26:18 `q004. The primes through 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. Twin primes are consecutive odd numbers which are both prime. Are there any twin primes in the set of primes through 40?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Twin primes through the number 40 would be = 3 and 5, 5and 7, 11and 13, 17and19, 29and 31.
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14:30:13 `q006. We can prove that 89 is prime as follows: 89 is odd and is hence not divisible by 2. If we divide 89 by 3 we get a remainder of 2 so 89 is not divisible by 3. Since 89 is not divisible by 2 it cannot be divisible by 4 (if we could divide it evenly by 4 then, since 2 goes into 4 we could divide evenly by 2; but as we just saw we can't do that). Since 89 doesn't end in 0 or 5 it isn't divisible by 5. If we divide 89 by 7 we get a remainder of 5 so 89 is divisible by 7. Since 89 isn't divisible by 2 it isn't divisible by 8, and since it isn't this will by 3 it isn't divisible by 9. At this point it might seem like we have a long way to go--lots more numbers to before get to 89. However it's not as bad as it might seem. For example once we get past 44 we're more than halfway to 89 so the result of any division would be less than 2, so there's no way it could be a whole number. So nothing greater than 44 is a candidate. We can in fact to even better than that. If we went even as far as dividing by 10, the quotient must be less than 10: since 10 * 10 is greater than 89, it follows that 89 / 10 must be less than 10. So if 10 or, more to the point, anything greater than 10 was going to divide 89 evenly, in the result would be one of the numbers we have already unsuccessfully tried. It follows that after trying without success to divide 89 by all the numbers through 9, which we didn't really have to try anyway because 9 is divisible by 3 which we already checked, we are sure that 89 has to be prime. What is the largest number you would have to divide by to see whether 119 is prime?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. to see if 119 is prime, the largest number you would have to divide by is 3
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14:31:13 After we get to 7, we don't have to try 8, 9 or 10 because we've already checked numbers that divide those numbers. The next number we might actually consider trying is 11. However 11 * 11 is greater than 119, so any quotient we get from 11 on would be less than 11, and we would already have eliminated the possibility that any number less than 11 is a divisor. So 7 is the largest number we would have to try.
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RESPONSE --> ok. 7 is the largest number to divide by to find out if 119 is a prime number.
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14:32:11 `q007. Precisely what numbers would we have to try in order to determine whether 119 is prime?
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RESPONSE --> number to try to see if the number 119 is prime are 1,3,5,7.
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14:33:12 We would have to check whether 119 was divisible by 2, by 3, by 5 and by 7. We wouldn't have to check 4 or 6 because both are divisible by 2, which we would have already checked.
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RESPONSE --> to find out if 119 is prime you wouldn't have to check 4 or 6 because they are both divisible by 2, and we already checked 2.
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14:36:18 `q008. Is 119 prime?
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RESPONSE --> 119 is not a prime number.
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14:36:42 119 isn't divisible by 2 because 119 is odd. 119 isn't divisible by 3 because 120 is. 119 isn't divisible by 5 because 119 doesn't end in 5. 119 is, however, divisible by 7, as you can easily verify.
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RESPONSE --> 119 is divisible by 7
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14:47:06 `q009. We can 'break down' the number 54 into the product 2 * 27, which can further be broken down to give us 2 * 3 * 9, which can be broken down one more step to give us 2 * 3 * 3 * 3. We could have broken down 54 into different way as 6 * 9, which could have been broken down into 6 * 3 * 3, which can be broken down one more step to give us 2 * 3 * 3 * 3. No matter how we break 54 down into factors, the process ends with a single factor 2 and 3 repetitions of the factor 3. Break down each of the following in this manner, until it is not possible to break it down any further: 63, 36, and 58.
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RESPONSE --> Breaking down the following number 63= 7*9 broke into = 7*3*3=63 36= 4*9 = 2*2*3*3=36 58= 2*29= 2*29
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14:49:06 `q010. The results that we obtained in the preceding exercise are called the 'prime factorizations' of the given numbers. We have broken the numbers down until we are left with just prime numbers. Why is it that this process always ends with prime numbers?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. When you break down numbers you always end up with prime numbers becasue those are the numbers divisible by that number and one. You can not break them down any farther.
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14:49:16 We break the numbers down until they can't be broken down any further. If any number in the product is not prime, then it has more than two factors, which means it has to be divisible by something other than itself or 1. In that case we would have to divide it by that number or some other. So the process cannot end until all the factors are prime.
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RESPONSE --> ok
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14:54:16 `q011. Find the prime factorization of 819, then list all the factors of 819.
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RESPONSE --> The prime factorization of 819 is first 9*91=819, broke down to 3*3*7*13.
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14:54:29 We start by dividing by 3. We get 819 = 3 * 273 = 3 * 3 * 91. Since 91 isn't divisible by 3, we need to 5 and 7, after which the next prime is 11 and which we will not need to try since 11 * 11 > 91. 91 isn't divisible by 5 since it doesn't end in 0 or 5, but it is divisible by 7 with quotient 13. So 91 = 7 * 13. Thus we have 819 = 3 * 3 * 91 = 3 * 3 * 7 * 13. In addition to 1 and 819, the factors of 819 will include 3, 7, 13, all of the prime factors, 3 * 3 = 9, 3 * 7 = 21, 3 * 13 = 39 and 7 * 13 = 91, all of the possible products of two of the prime factors; and 3 * 3 * 7 = 63, 3 * 3 * 13 = 117, and 3 * 7 * 13 = 273, all of the possible products of three of the prime factors.
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RESPONSE --> ok
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15:01:09 `q012. List all the factors of 168.
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RESPONSE --> factors of 168 are 2,3,6,7,8,12,14,21,24,28,42,56 snd 84.
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15:01:33 First we find the prime factorization: 168 = 2 * 84 = 2 * 2 * 42 = 2 * 2 * 2 * 21 = 2 * 2 * 2 * 3 * 7. This number has several factors: We have the prime factors themselves: 2, 3, 7. We have the products of exactly 2 of the prime factors: 2 * 2 = 4, 2 * 3 = 6, 2 * 7 = 14, 3 * 7 = 21. We have products of exactly 3 of the prime factors: 2 * 2 * 2 = 8; 2 * 2 * 3 = 12; 2 * 2 * 7 = 28; 2 * 3 * 7 = 42. We have products of exactly 4 of the prime factors: 2 * 2 * 2 * 3 = 24; 2 * 2 * 2 * 7 = 56; and 2 * 2 * 3 * 7 = 84. And we finally have 168 itself (the product of all 5 prime factors) and 1. Note that using exponential notation we can write the prime factorizations of 168 as 2^3 * 3^1 * 7^1. We have 2 as a factor 3 times, and 3 and 7 each one time. If we add 1 to the power of each prime factor we get 3 + 1 = 4, 1 + 1 = 2 and 1 + 1 = 2. If we then multiply the resulting numbers together we get 4 * 2 * 2 = 16. Note that there are 16 factors of 168. This process always works: if we add 1 to the power of each prime factor then multiply the results we get the total number of factors of the original number.
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RESPONSE --> ok
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FԐwڼPyxh Student Name: assignment #025
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16:04:08 `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 --> the prime factorizations of 60 and 105, both have the prime numbers 3 and 5 used one time in common. the smalles prime for both is 3.
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16:09:52 `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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. prime factorization of 84 = 7*12= 7*3*4 = 7*3*2*2 which is 2*2*3*7 = 84. 126= 3*42 = 3*3*14= 3*3*2*7 which is 2*3*3*7
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16:10:12 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
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16:10:20 `q003. Find the greatest common divisor and least common multiple of 504 and 378.
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RESPONSE -->
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ғQ˰mٹܷ Student Name: assignment #026
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16:16:43 `q001. There are six questions in this assignment. We defined an operation as follows: x * y (mod 4) = remainder when x * y is divided by 4. Find 3 * 9 (mod 4); 7 * 12 (mod 4) and 11 * 13 (mod 4).
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RESPONSE --> I am not sure.
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16:17:40 3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see that 3 * 9 (mod 4) = 3. 7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we see that 7 * 12 (mod 4) = 0. 11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder 3,we see that 11 * 13 (mod 4) = 3.
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RESPONSE --> ok
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16:23:31 `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x * y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. * mod 5 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 0 3 3 3 0 3 0 2 2 4 0 4 3 2 1
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16:25:16 `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x is a solution to this equation?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. 3x + 7 = 9 3x = 9-7 3x = 2 x= 2-3 x=1
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ݨ笩x}` Student Name: assignment #024
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15:36:30 `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 --> two prime numbers adding to 28 could be 5+23=28
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15:36:39 28 can be expressed as 5 + 23, both of which are prime.
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RESPONSE --> yes
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15:40:04 `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= 11+31 76= 29+47
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15:41:11 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 --> 42 can equal 23+19, 13+29,11+31, or 5+37 76 = 73+3,71+5,59+17,23+23,29+47 there can be more than one answer.
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15:43:04 `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 proper factors of 18 are 1,2,3,6,9 added all together they = 21. That is more than 18.
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15:43:12 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 --> ok
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15:47:15 `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, = 16. Therefore this number is abundant. 26= 1,2,13. = 16. This number is deficient. 16= 1,2,4,8 = 15. This number is deficient. 6= 1,2,3 = 6. This number is perfect.
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15:47:25 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 --> yes
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15:55:29 `q005. There is a perfect number between 20 and 30. Find it.
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RESPONSE --> a perfect number between 20 and 30 is. 20= 1,2,4,5,10 - 22 21= 1,3,7= 11 22= 1,2,11 = 14 23= PRIME 24=1,2,3,4,6,8,12 = 36 25= 1,5, = 6 26= 1,2,13 = 16 27= 1,3,9, = 13 28= 1,2,4,7,14 = 28 The perfect number is 28
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15:55:35 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 --> yes
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15:57:05 `q006. Why can't a prime number be perfect?
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RESPONSE --> a prime number can't be perfect because it is only divisible by 1 and itself.
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15:57:13 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 --> yes
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15:57:49 `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 --> no
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15:58:21 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. ok
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