course Mth 151 k~vOFassignment #020
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20:21:22 query 4.3.6 number following base-six 555
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RESPONSE --> The only numbers that exist in base-six are 0,1,2,3,4,5. The number before 555 is 554, because all of these numbers exist in base-six. The next number after 555 that contains nothing but numbers that exist in base-six is 1000. confidence assessment: 3
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20:21:31 ** 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 --> OK self critique assessment: 3
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20:23:10 query 4.3.20 34432 base five
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RESPONSE --> 34432 base five (3 * 5^4) + (4 * 5^3) + (4 * 5^2) + (3 * 5^1) + (2* 5^0) 1875 + 500 + 100 + 15 + 2 = 2492 confidence assessment: 3
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20:23:15 **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 --> OK self critique assessment: 3
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20:25:12 Explain how you use the calculator shortcut to get the given number.
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RESPONSE --> The calculator shortcut for the previous problem would be: 2 + 5(3+5(4+5(4+3*5) Working backwards, you simply multiply 3 by 5, add 4, multiply by 5, add 4, multiply by 5, add 3, multiply by 5, and add 2. confidence assessment: 3
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20:25:16 ** 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 --> OK self critique assessment: 3
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20:29:32 query 4.3.40 11028 decimal to base 4
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RESPONSE --> The base 4 place values are 1, 4, 16, 64, 256, 1024, 4096 11028 / 4096= 2 remainder 2836 2836 / 1024 = 2 remainder 788 788 / 256 = 3 remainder 20 we don't need to divide by 64, so we use a 0 20 / 16 = 1 remainder 4 4 / 4 = 1 remainder 0 0 / 1 = 0 2230110 in base four confidence assessment: 3
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20:29:40 ** 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 --> OK self critique assessment: 3
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20:30:29 query 4.3.51 DC in base 16 to binary
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RESPONSE --> D stands for 13, which is 1101 in binary C stands for 12, which is 1100 in binary 11011100 in binary confidence assessment: 3
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20:30:38 ** 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 --> OK self critique assessment: 3
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20:32:16 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 --> It is sometimes even and sometimes odd. For example, 82 base nine (8 * 9^1) + (2 * 9^0) = 74, is even But 70 base nine (7 * 9^1) + (0 * 9^0) = 63, is odd confidence assessment: 3
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20:32:29 ** 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 --> OK self critique assessment: 3
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