course Mth 151 ԎwKxp\i聣Z_܄assignment #020
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16:05:20 query 4.3.6 number following base-six 555
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RESPONSE --> 556. Would it be the next counting number? confidence assessment: 1
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16:14:45 ** COMMON ERROR: 556. INSTRUCTOR COMMENT: The digit 6 does not exist in base six for the same reason as there is no single digit for 10 in base 10 (digits stop at 9; in general there is no digit corresponding to the base). CORRECT SOLUTION: 555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get 5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to 5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have 5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have 6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is 6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0. So the number following 555{base 6} is 1000{base 6}. The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000. **
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RESPONSE --> Okay, it would be just like adding 1 to 9, it would be ten which moves it up to tens instead of still being ones. self critique assessment: 2
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16:18:41 query 4.3.20 34432 base five
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RESPONSE --> 3*5^4 + 4*5^3 + 4*5^2 + 3*5^1 + 2*5^0 1875 + 500 + 100 + 15 + 2 = 2492 confidence assessment: 2
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16:19:21 **34432 base five means 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. 5^2 = 625, 5^3 = 125, 5^2 = 25, 5^1 = 5 and 5^0 = 1 so 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492. **
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RESPONSE --> Work through your powers then mutiply out and then add them all. self critique assessment: 2
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16:24:25 Explain how you use the calculator shortcut to get the given number.
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RESPONSE --> (((3*5+4) * 5+4) * 5+3) * 5 + 2). I have a hard time with the ( ) and making sure they are in right. confidence assessment: 2
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16:26:53 ** 2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5s to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. **
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RESPONSE --> I'm sorry, but its easier for me to do it by hand. I'm always afraid that I will put it in wrong so I doubt my answer. self critique assessment: 2
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16:31:58 query 4.3.40 11028 decimal to base 4
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RESPONSE --> 2757 When you begin dividining to get the answer 4 will go into 11028 2757 times evenly leaving no remainder. confidence assessment: 2
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16:35:08 ** 4^0 = 1 4^1 = 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 (4*7 = 16386, which is larger than the given 11028) So to build up 11028 we need 2 * 4^6 = 8192, leaving 2836. 2 * 4^5 = 2048, leaving 788. 3 * 4^4 = 768, leaving 20. 0 * 4^3, because we need only 20, which is less than 64. 1 * 4^2 = 16, leaving 4. 1 * 4^1 = 4, leaving 0.0 * 4^1. Thus our number is 2230110 base 4.
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RESPONSE --> I was doing it like example 4 on pg. 181. They began with the number and divided it by the base but I didn't get anything close to the correct answer. I will do it how you subtract it and go to the next power self critique assessment: 2
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16:43:19 query 4.3.51 DC in base 16 to binary
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RESPONSE --> 11011100base2 confidence assessment: 2
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16:44:09 ** C stands for decimal 12, which in binary is 1100. D stands for decimal 13, which in binary is 1101. Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C. Note that this method works only when one base is a power of the other.**
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RESPONSE --> I used the chart in the book to convert the number self critique assessment: 2
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16:47:56 Is a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number?
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RESPONSE --> I would say even because the number ends with an even number????? confidence assessment: 2
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16:51:25 ** You can investigate this question by trying a variety of examples. For example, 82nine = 8 * 9^1 + 2 * 9^0 = 72 + 2 = 74 in decimal. This is even because you are adding two even multiples of the odd numbers 9^1 = 9 and 9^0 = 1. You have to use an even multiple of 9^0 = 1 because the number ends with an even number. But you don't have to use an even multiple of 9^1 = 9. So we try something like 72nine = 7 * 9^1 + 2 * 9^0 = 63 + 2 = 65 in decimal and we see that the result is odd. The key is that in base nine, the powers of nine are always odd numbers. So when converting you get from a base-9 number with two even digits a number which is even * odd + even = even, while from a base-9 number with an odd digit followed by an even digit you get odd * odd + even = odd; many other possible combinations show the same thing but this is one of the simpler examples that shows why we get odd for some base-9 numbers, and even for others. For a couple of specifics, 70 in base 9 is 63 in base ten, and is odd. However 770 in base 9 is even. **
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RESPONSE --> I understand that you get can get odd and even depending on the the entire row of numbers that you use with the base. self critique assessment: 2
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