course Mth 151
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
020. `query 20
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `qquery 4.3.6 number following base-six 555
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Your solution:
I don’t quite understand this and the book that I can find talks about it.
Confidence Assessment:
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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. **
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Self-critique (if necessary):
So basically I am just expanding the number in powers of 6?
Right. That's the meaning of 'base-six 555'.
Self-critique Rating:
Question: **** `qquery 4.3.20 34432 base five
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Your solution:
34432 five = (((3*5+4)*5+4)*5+3)*5+2 = 492
When adding it in my calculator I forgot to put in the *5+4 the second time making it 2492.
Confidence Assessment:
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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. **
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Self-critique (if necessary):
Ok I am now totally confused. The book said to use the calculator shortcut method in which I did exactly how the book said to do it so why am I off by 2000? I figured it out I put the equation in my calculator wrong.
Self-critique Rating:
Question: **** `qExplain how you use the calculator shortcut to get the given number.
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Your solution:
I put 2 less than the number of digits in parenthesis. I then start with the number on the left and multiply it with the base then adding the next digit and so on until the last number not multiplying it by the base.
Confidence Assessment:
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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. **
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Self-critique (if necessary):
I set mine up like the book which is backwards from yours.
Self-critique Rating:
Question: **** `qquery 4.3.40 11028 decimal to base 4
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Your solution:
2*4^6=8192
2*4^5=2048
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11028-10240 = 788
3*4^4=768
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788-768= 20
1*4^2=16
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20-16= 4
4*4^0= 4
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4-4=0
2*4^6+2*4^5+3*4^4+1*4^2+4*4^0=11028
Confidence Assessment:
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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.
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Self-critique (if necessary):
I forgot to put 0*4^3 because I wasn’t using it.
You want to do that, of course, but it's a minor error. You did very well in your explanation.
Self-critique Rating:
Question: **** `qquery 4.3.51 DC in base 16 to binary
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Your solution:
DC=11011100
Confidence Assessment:
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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.**
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Self-critique (if necessary):
I understand
Self-critique Rating:
Question: **** `qIs 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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Your solution:
9^1=9
9^2=81
9^3=729
9^4=6561
9^5=59049
9^6=531441
It looks like it is always odd.
Confidence Assessment:
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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. **
I was looking at it wrong.
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This looks good. See my notes. Let me know if you have any questions.