#$&* course Mth 151 Your videos are labeled wrong for Chapters 4 and 5. I was watching the videos for Chapter 5 and trying to do the assignments for Chapter 4. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: In base 10, 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 3 * 16 + 2 * 4 + 1 * 1 = 48 + 8 + 1 = 57. STUDENT COMMENT: I am not understanding this. INSTRUCTOR RESPONSE statement 1: 321{base 4} means 3 * 4^2 + 2 * 4^1 + 1 * 4^0. statement 2: 3 * 4^2 + 2 * 4^1 + 1 * 4^0 = 57. What is it you do and do not understand about the above two statements? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Your question asked us to do the problem in base 10, but you did your solution in base 4. I don’t understand that. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: 213{base 4} means 2 * 4^2 + 1 * 4^1 + 3 * 4^0 = 2 * 16 + 1 * 4 + 3 * 1 = 32 + 4 + 3 = 39. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q003. Suppose we had a number expressed in the form 6 * 4^2 + 7 * 4^1 + 3 * 4^0. In base 4 every term needs to be expressed in the highest possible power of 4. This is not the case for the given number, since for example the coefficient 7 can be expressed as 1 * 4^1 + 3 * 4^0. How would the number 6 * 4^2 + 7 * 4^1 + 3 * 4^0 be expressed without using any coefficients greater than 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Can’t figure this one out. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 * 4^0 = (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): I can’t figure this out, because I’m not sure what a coefficient is. Even after looking up the definition, I’m not sure. When you start off writing 7 as 4 * 4˄2, isn’t that first four a coefficient? If so, it is more than 3. I wish you had done a video on this kind of problem. ------------------------------------------------ Self-critique Rating: 0
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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): Looking ahead and still cannot figure out what I’m supposed to be doing. I can follow you problem, I just cannot figure out what the problem is asking me to do. So, I guess this means when we add 1 to the number, it turns it into 2000.
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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): WOW! I understood this one and got it right. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. Find the base-10 equivalent of the number 322{base 4}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 * 4˄2 + 2 * 4˄1 + 2 * 4˄0 48 + 8 + 1 = 57 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q007. Find the base-4 equivalent of the number 487. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 * 4˄4 + 3 * 4˄3