course Mth 151 zyStudent Name:
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17:28:16 `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 --> 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 (1,2,4,8,16) 17 (1,17) 18 (1,2,3,6,9,18) 19 (1,19) 20 (1,2,4,5,10,20)
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17:28:22 The divisors of 2 are 1 and 2. The divisors of 3 are 1 and 3. The divisors of 4 are 1, 2 and 4. The divisors of 5 are 1 and 5. The divisors of 6 are 1, 2, 3 and 6. The divisors of 7 are 1 and 7. The divisors of 8 are 1, 2, 4, hence 8. The divisors of 9 are 1, 3 and 9. The divisors of 10 are 1, 2, 5 and 10. The divisors of 11 are 1 and 11. The divisors of 12 are 1, 2, 3, 4, 6, 12. The divisors of 13 are 1 and 13. The divisors of 14 are 1, 2, 7 and 14. The divisors of 15 are 1, 3, 5 and 15. The divisors of 16 are 1, 2, 4, 8 and 16. The divisors of 17 are 1 and 17. The divisors of 18 are 1, 2, 3, 6, 9 and 18. The divisors of 19 are 1 and 19. The divisors of 20 are 1, 2, 4, 5, 10 and 20.
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RESPONSE --> k
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17:28:58 `q002. Some of the numbers you listed have exactly two divisors. Which are these?
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RESPONSE --> 2,3,5,7,11,13,17,19
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17:29:04 The numbers with exactly two divisors are 2, with divisors 1 and 2, 3, with divisors 1 and 3, 5, with divisors 1 and 5, 7, with divisors 1 and 7, 11, with divisors 1 and 11, 13, with divisors 1 and 13, 17, with divisors 1 and 17 and 19, with divisors 1 and 19.
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RESPONSE --> k
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17:29:36 `q003. These numbers with exactly two divisors are called prime numbers. List the prime numbers between 21 and 40.
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RESPONSE --> 23, 29, 31, 37
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17:29:41 Every counting number except 1 has at least 2 divisors, since every counting number is divisible by itself and by 1. Therefore if a counting number is divisible by any number other than itself and 1 it has more than 2 divisors and is therefore prime. {}Since 21 is divisible by 3, it has more than 2 divisors and is not prime. Since 22 is divisible by 2, it has more than 2 divisors and is not prime. Since this will be the case for all even numbers, we will not consider any more even numbers as candidates for prime numbers. Since 23 is not divisible by any number except itself and 1, it has exactly 2 divisors and is prime. Since 25 is divisible by 5, it has more than 2 divisors and is not prime. Since 27 is divisible by 3, it has more than 2 divisors and is not prime. Since 29 is not divisible by any number except itself and 1, it has exactly 2 divisors and is prime. Since 31 is not divisible by any number except itself and 1, it has exactly 2 divisors and is prime. Since 33 is divisible by 3, it has more than 2 divisors and is not prime. Since 35 is divisible by 5, it has more than 2 divisors and is not prime. Since 37 is not divisible by any number except itself and 1, it has exactly 2 divisors and is prime. Since 39 is divisible by 3, it has more than 2 divisors and is not prime. The primes between 21 and 40 are therefore 23, 29, 31 and 37.
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RESPONSE --> k
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17:30:18 `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 --> 2 and 3
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17:30:41 17 and 19 are consecutive odd numbers, and both are prime. The same is true of 5 and 7, and of 11 and 13.
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RESPONSE --> wow, i'm dumb. i didn't fully read that question
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17:33:50 `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 --> 53
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17:34:08 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 --> right, right
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17:34:30 `q007. Precisely what numbers would we have to try in order to determine whether 119 is prime?
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RESPONSE --> 2,3,5,7
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17:34:40 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 --> k
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17:34:58 `q008. Is 119 prime?
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RESPONSE --> yes it is
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17:35:09 119 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 --> right
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17:37:42 `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 --> 63(3*21, 3*3*7), 36(6*6, 2*2*3*3), 58(2*29)
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17:37:49 We see that 63 = 9 * 7 = 3 * 3 * 7, 36 = 9 * 4 = 3 * 3 * 4 = 3 * 3 * 2 * 2, which we rearrange in increasing order of factors as 2 * 2 * 3 * 3, and 58 = 2 * 29.
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RESPONSE --> k
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17:39:31 `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 --> cause everything is divisble by a prime number, some might take longer than others but the fact remains that every counting numbers is divisible by a prime so if you break it all the way down eventually it is divisible by a prime
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17:39:37 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 --> k
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17:41:03 `q011. Find the prime factorization of 819, then list all the factors of 819.
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RESPONSE --> (3*273), (3*91), so 3, 3, 91, 273
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17:41:19 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 --> forgot a couple
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17:45:09 `q012. List all the factors of 168.
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RESPONSE --> 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168
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17:45:15 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 --> k
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һߗߝ Student Name: assignment #024
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18:09:11 `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 --> 23+5
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18:09:18 28 can be expressed as 5 + 23, both of which are prime.
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RESPONSE --> right
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18:10:27 `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 --> 19+23, 53+23
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18:10:35 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 --> k
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18:47:32 `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 --> 1, 2, 9, so less than which would make itabundant
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18:48:02 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 --> darn, i left 6 out, so that would be deficient
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18:55:00 `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 deficient 26. 1, 2, 13=16 abundant 16. 1, 2, 4, 9=16, perfect 6. 1, 2, 3,=6 perfect
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18:55:20 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 --> K
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18:56:04 `q005. There is a perfect number between 20 and 30. Find it.
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RESPONSE --> 28
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18:56:08 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 --> k
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18:56:50 `q006. Why can't a prime number be perfect?
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RESPONSE --> cause it has to have proper divisors
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18:57:16 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 --> k
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18:59:07 `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 cause 63 isn't prime, it's divisable by 3
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18:59:21 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 --> k
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19:03:12 `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 --> yes for 3 cause 2^3-1=7 yes for 5 cause =31 yes for 7 cause =127
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19:03:22 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 --> right
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cqͭۺÒnR Student Name: assignment #025
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19:28:48 `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 --> 3 and 5, i would guess 5 and 7
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19:29:15 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 --> gotcha
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20:02:04 `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 --> 84) 2^2*3^1*7*1 126)2^1*3^2*7^1 that would be 7 LCM) 2^2 * 3^2 * 7^1=252
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20:02:09 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 --> k
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20:07:16 `q003. Find the greatest common divisor and least common multiple of 504 and 378.
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RESPONSE --> 504( 2^3 * 3^2 * 7^1) 378 ( 2^1 * 3^2 *7^1) 7 again for GCD 2^3 * 3^2 * 7^1=504
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20:08:29 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 --> i didn't combine
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ᕗGJƊJx Student Name: assignment #026
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20:33:15 `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 --> 3*9(3) 7 * 12(0) 11 * 13 (3)
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20:33:20 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 --> k
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20:36:50 `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set {0, 1, 2, 3}. Determine which of the properties, including commutativity, associative, identity, inverse and closure properties, are properties of this operation.
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RESPONSE --> mod 4\ 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 well, zero has the identity property, everything else is associative
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20:37:12 Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0. Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4 (e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we can see that 1 is the identity for this operation. Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2, respectively. Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1, respectively. The table for this operation is therefore * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the elements of the set to give us the identity, the operation on this set does not have the inverse property. We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we could in any event have concluded from the fact that multiplication is commutative so that the product we get before calculating the remainder is independent of the order of the two numbers. In a similar matter we can reason that the operation is associative. The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the set {0, 1, 2, 3}.
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RESPONSE --> k
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20:40:10 `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 --> mod 5/ 1 2 3 4 5 1 1 2 3 4 5 2 2 4 1 3 0 3 3 1 4 2 0 4 4 3 2 1 0 5 0 0 0 0 0 identity and inverse
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20:40:46 First we might wish to do a couple of example calculations to get familiar with the operation. For example: 2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1. 3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2. 2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3. The table is * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed. This operation has identity 1, as we can see from the row and the column across from and beneath 1. We easily see from the table that the identity appears exactly once in each row and in each column, which assures us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own inverse. The associativity and commutativity of the operation follow from the associative and commutative properties of multiplication on real numbers, as discussed in the preceding problem.
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RESPONSE --> ok
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20:41:47 `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 --> i'm not sure
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20:44:12 3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0. If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0 when dividing -2 by 5). If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1. If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4. If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2. If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0. Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5).
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RESPONSE --> i didn't understand this one
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22:00:03 `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6, 7, 8, 9 is also a solution. Which one is it?
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RESPONSE --> 8
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22:00:28 We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution. We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens. In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n.
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RESPONSE --> k
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22:00:45 `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem?
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RESPONSE --> ?
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22:01:05 We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5 solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24.
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RESPONSE --> k
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v{ԚÝq颍~ɜs Student Name: assignment #028
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22:28:07 `q001. Note that there are 19 questions in this section. Solve the equation 2 x + 7 = 21.
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RESPONSE --> 2x=14 x=7
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22:28:14 To solve a linear equation we can use the basic properties of equivalents and equality. Specifically we can simplify either or both sides of an equation, we can add or subtract the same quantity from both sides of the equation, or we can multiply or divide both sides of the equation by the same quantity. To solve the present equation we need to obtain any equivalent equation with just x on one side or the other. We proceed as follows: Starting with the original equation 2 x + 7 = 21, we first subtract 7 from both sides to obtain 2x + 7 - 7 = 21 - 7. We simplify both sides to obtain 2x = 14, then we divide both sides of the equation by 2 to get 2x / 2 = 14 / 2. Simplifying both sides we have x = 7.
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RESPONSE --> k
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22:29:07 `q002. Solve the equation 2 (x + 7) = 30.
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RESPONSE --> 2x+14=30 2x=16 x=8
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22:29:11 Starting with 2 (x+7) = 30 we first simplify the left-hand side using the distributive law of multiplication over addition. We obtain 2x + 14 = 30. We subtract 14 from both sides to obtain 2x = 16, then we divide both sides by 2 to obtain{}{}2x / 2 = 16 / 2 or after simplifying x = 8.
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RESPONSE --> k
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22:29:36 `q003. Using x for the variable, how do we write the expression 'double a number plus five'?
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RESPONSE --> 2x+5
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22:29:40 If x stands for the number, then double the number is 2x, and double the number plus 5 is therefore 2x + 5.
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RESPONSE --> K
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22:29:58 `q004. Using x for the variable, how do we express the statement 'double a number plus five is 19'?
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RESPONSE --> 2x+5=19
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22:30:03 As seen in the preceding question, 'double a number plus five' can be expressed as 2 x + 5, so the statement 'double a number plus five is 19' can be written as the equation 2x + 5 = 19.
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RESPONSE --> k
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22:30:48 `q005. Using x for the length of the side of a square, how do we express the perimeter of the square?
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RESPONSE --> 4x=perimeter
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22:30:58 The perimeter of a figure is the distance around it. A square consists of 4 equal sides. If each side has length x, then the distance around it is 4 * x, or just 4 x.
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RESPONSE --> k
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22:31:31 `q006. Using an appropriate variable for the length of the side of a square, how do we express the area of a square in terms of the length of its side?
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RESPONSE --> x*x
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22:31:39 The area of a square is the product of the lengths of its sides. If x stands for the length of a side, then the area of the square is x * x or x^2.
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RESPONSE --> k
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22:32:03 `q007. Using an appropriate variable how do we express the perimeter of a rectangle whose length is double its width?
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RESPONSE --> 2l * w
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22:32:14 The perimeter is the sum of the lengths of the sides of the rectangle. Since we are given the relationship between the lengths of the sides, we first express this relationship in symbols. If we let x stand for the width of the rectangle, then the length of the rectangle is 2x. Since both the length and the width occur twice as we move around the perimeter of the rectangle, it follows that the perimeter is 2 * length + 2 * width = 2 * 2x + 2 * x. This expression simplifies to 4 x + 2 x, which further simplifies to 6x.
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RESPONSE --> oh ok
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22:34:49 `q008. Using an appropriate variable, how do we express the area of a rectangle whose length is double its width?
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RESPONSE --> 2lw
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22:35:01 The area of a rectangle is the product of its length then width. Since we are given the relationship between length in width, we begin by expressing the length and width of the rectangle in terms of a symbol. If we let x stand for the width of the rectangle, then 2 x stands for its length. Now since the area is the product of length and width, we see that the area must be x * 2 x, or 2 x^2.
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RESPONSE --> k
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22:35:45 `q009. Express in terms of an appropriate variable the statement that the area of a rectangle whose length is double its width is 72. Solve the resulting equation to determine the dimensions of the rectangle.
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RESPONSE --> 2x * w= 72
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22:35:54 The area of a rectangle whose length is double its width was seen in the preceding problem to be 2 x^2, where x is the width. If the area is 72, we see that 2 x^2 = 72. We solve this equation as follows. Starting with 2 x^2 = 72 we divide both sides by 2 to get x^2 = 36. We then take the square root of both sides of the equation to obtain x = `sqrt(36), or x = 6. Note that the equation x^2 = 36 actually has two solutions, x = 6 and x = -6. However the length of a side cannot be negative so we chose the positive solution.
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RESPONSE --> k
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