#$&* course Mth 151 4/13/2012 3:43PM Question: `q001. There are 5 questions in this set.
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Given Solution: In base 10, 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 3 * 16 + 2 * 4 + 1 * 1 = 48 + 8 + 1 = 57. STUDENT COMMENT: I am not understanding this. INSTRUCTOR RESPONSE statement 1: 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0. statement 2: 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57. What is it you do and do not understand about the above two statements? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. What would the number 213{base 4} be in base 10 notation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 213 {base 4} = 2 * 4^2 + 1 * 4^1 + 3 * 4^0, in base 10 it would work out to 2 * 16 + 1 * 4 + 3 * 1 = 32 + 4 + 3 = 39. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 213{base 4} means 2 * 4^2 + 1 * 4^1 + 3 * 4^0 = 2 * 16 + 1 * 4 + 3 * 1 = 32 + 4 + 3 = 39. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. Suppose we had a number expressed in the form 6 * 4^2 + 7 * 4^1 + 3 * 4^0. This number isn't quite in the form needs to be if it is to be expressed in base 4. This is because we have the numbers 6 and 5, which exceed 4. How would this number be expressed without using any numbers 4 or greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 * 4^2 + 7 can be changed to 4 * 4^2 + 2 * 4^2 and 7 * 4^1 can be changed to 4 * 4^1 + 3 * 4^1, because 4 + 2 = 6 and 4 + 3 = 7. So it would now be 4 * 4^2 + 2 * 4^2 + 4 * 4^1 + 3 * 4^1 + 3 * 4^0. 4 * 4^2 equals 4^3 and 4 * 4^1 equals 4^2. Now the it is 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 which equals 1333. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 7 = 4 + 3 so 7 * 4^1 can be written as 4 * 4^1 + 3 * 4^1 = 4^2 + 3 * 4^1 Since 6 = 4 + 2, we have 6 * 4^2 = 4 * 4^2 + 2 * 4^2. Since 4 * 4^2 = 4^3, this is 4^3 + 2 * 4^2. Thus 6 * 4^2 + 7 * 4^1 + 3 * 4^1 = (4 * 4^2 + 2 * 4^2) + (4 * 4^1 + 3 * 4^1) + 3 * 4^0 =4^3 + 2 * 4^2 + 4^2 + 3 * 4^1 + 3 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0. This number would then be 1333 {base 4}. STUDENT COMMENT I understand the answer, but not the first paragraph of the explanation. INSTRUCTOR RESPONSE Here is an expanded version of the first line: 7 * 4^1 = (4 + 3) * 4^1 = 4 * 4^1 + 3 * 4^1. Since 4 * 4^1 = 4^2, it follows that 7 * 4^1 = 4^2 + 3 * 4^1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. What would happen to the number 1333{base 4} if we added 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 equals 1 * 4^0, and that added to 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 4 * 4^0. The 4 * 4^0 would change to 4^1, which make 3 * 4^1, 4 * 4^1, and that would change to 4^2, and that would make 3 * 4^2, 4^3 which would make 1 * 4^3, 2 * 4^3. So the equation is 2 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0, which equals 2000. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since 1 = 1 * 4^0, Adding one to 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 would give us 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 3 * 4^0 + 1 * 4^0 = 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 4 * 4^0. But 4 * 4^0 = 4^1, so we would have 1 * 4^3 + 3 * 4^2 + 3 * 4^1 + 1 * 4^1 + 0 * 4^0 = 1 * 4^3 + 3 * 4^2 + 4 * 4^1 + 0 * 4^0 . But 4 * 4^1 = 4^2, so we would have 1 * 4^3 + 3 * 4^2 + 1 * 4^2 + 0 * 4^1 + 0 * 4^0 = 1 * 4^3 + 4 * 4^2 + 0 * 4^1 + 0 * 4^0 . But 4 * 4^2 = 4^3, so we would have 1 * 4^3 + 1 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0 = 2 * 4^3 + 0 * 4^2 + 0 * 4^1 + 0 * 4^0. We thus have the number 2000{base 4}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. How would the decimal number 659 be expressed in base 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started with figure out the highest power of 4 I would need, I used 4^4 since it is the highest number that doesn’t go past 659. 4^4 equals 256, and I need to find the highest multiple of it that doesn’t go over 659, which would be 2. that equals 512 and leaves 147 of the 659 left, so next I found the highest multiple of 4^3 or 64 that doesn’t go over 147, which is also 2. that equals 128 and leaves me with 19. Then I found the highest multiple of 4^2 or 16, which is 1. That leaves me with 3, and then I found the highest multiple of 4^0 because 4^1 is already higher than 3. That would be 3 and make it 2 * 4^4 + 2 * 4^3 + 1 * 4^2 + 0 * 4^1 + 3 * 4^0. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We need to express 659 in terms of multiples powers of 4, with the multiple not exceeding 3. The powers of 4 are 4^0 = 4, 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024. We could continue to higher powers of 4, but since 4^5 = 1024 already exceeds 659 we need not do any further. The highest power of 4 that doesn't exceed 659 is 4^4 = 256. So we will use the highest multiple of 256 that doesn't exceed 659. 2 * 256 = 512, and 3 * 256 exceeds 659, so we will use 2 * 256 = 2 * 4^4. This takes care of 512 of the 659, leaving us 147 to account for using lower powers of 4. We then account for as much of the remaining 147 using the next-lower power 4^3 = 64. Since 2 * 64 = 128 is less than 147 while 3 * 64 is greater than 147, we use 2 * 64 = 2 * 4^3. This accounts for 128 of the remaining 147, which now leaves us 19. The next-lower power of 4 is 4^2 = 16. We can use one 16 but not more, so we use 1 * 16 = 1 * 4^2. This will account for 16 of the remaining 19, leaving us 3. This 3 is accounted for by 3 * 4^0 = 3 * 1. Note that we didn't need 4^1 at all. So we see that 659 = 2 * 4^4 + 2 * 4^3 + 1 * 4^2 + 0 * 4^1 + 3 * 4^0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!