#$&* course Mth 151 4/15/2012 8:13PM 020. `query 20
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Given Solution: `a** COMMON ERROR: 556. INSTRUCTOR COMMENT: The digit 6 does not exist in base six for the same reason as there is no single digit for 10 in base 10 (digits stop at 9; in general there is no digit corresponding to the base). CORRECT SOLUTION: 555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get 5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to 5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have 5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have 6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is 6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0. So the number following 555{base 6} is 1000{base 6}. The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery 4.3.20 34432 base five YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 34432 = (((3 * 5 + 4) * 5 + 4) * 5 + 3) * 5 + 2 = 2492 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qExplain how you use the calculator shortcut to get the given number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I multiplied the first digit by 5 then added it to the second. Then multiplied that by 5 and the next digit, and kept doing that until I worked throw each digit. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5’s to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery 4.3.40 11028 decimal to base 4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started with finding what power up 4 to use, which is 4^6. Next I found what to multiply by 4^6 to get as close to 11028 without going over, which is 2. That left me with 2836 and I do the same thing before for 4^5, which is also 2. That left me with 788 and I do the same thing before for 4^4, which is 3. That left me with 20 and I do the same thing before for 4^2, 4^3 is already higher than 20, which is 1. That left me with 4 and I do the same thing before for 4^1, which is also 1. That left me with 0 so the last thing is 0 * 4^0. The whole thing is 2 * 4^6 + 2 * 4^5 + 3 * 4^4 + 0 * 4^3 + 1 * 4^2 + 1 * 4^1 + 0 * 4^0, which equals 22303110. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. STUDENT QUESTION I don’t understand how to get this answer, I looked at the answer and either way it’s not the same number. I always get above or below the 11028. Your answer says 2230110 but I thought you were trying to get 11028. INSTRUCTOR RESPONSE 2230110 in base four is equivalent to the decimal number 2 * 4^6 + 2 * 4^5 + 3 * 4^4 + 0 * 4^3 + 1 * 4^2 + 1 * 4^1 + 0 * 4^0, which if multiplied out and added is 11028 in decimal form. The given solution shows how to reason this out, starting with the decimal number 11028. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery 4.3.51 DC in base 16 to binary YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: D equals 1101 and C equals 1100 in binary so DC in binary is 11011100. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** C stands for decimal 12, which in binary is 1100. D stands for decimal 13, which in binary is 1101. Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C. Note that this method works only when one base is a power of the other.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qIs a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Base-9 numbers can sometimes be even and sometimes be odd. Like in 34 base-9 equals 31in decimal, while 35 base-9 equals 32 decimal. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!