assingment 19qa

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course mth151

12/7/14 9:35

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.

019. Place-value System with Other Bases

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Question: `q001. There are 7 questions in this set.

The calculations of the preceding qa were done in our standard base-10 place value system. We can do similar

calculations with bases other than 10.

For example, a base-4 calculation might involve the number 3 * 4^2 + 2 * 4^1 + 1 * 4^0. This number will be

expressed as 321{base 4}.

What would this number be in base 10?

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Your solution:

321{base 4} is 3*4^2 + 2*4^1 + 4*^0 = 3*16 +2*4 + 1*1 = 48 + 8 +1 =57

confidence rating #$&*: 3

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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?

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q002. What would the number 213{base 4} be in base 10 notation?

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Your solution:

213{base 4} is 2*4^2 + 1*4^1 + 3*4^0 = 2*16 + 1*4 + 3+1 = 32+4+1 = 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.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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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?

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Your solution:

6 * 4^2 = 1*4^3 + 2*4^2

7 * 4^1 = 1*4^2 +3*4^1

1*4^3 + 2*4^2 + 1*4^2 + 3*4^1 + 3*4^0

1*4^3 + 3*4^2 + 3*4^1 + 3*4^0 = 1333{base 4}

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.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q004. What would happen to the number 1333{base 4} if we added 1?

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Your solution:

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*4^1

4*4^0 = 1*4^1

1*4^3 + 3*4^2 + 3*4^1 + 1*4^1

1*4^3 + 3*4^2 + 4*4^1

4*4^1 = 1*4^2

1*4^3 + 3*4^2 + 1*4^2

1*4^3 + 4*4^2

4*4^2 = 1*4^3

1*4^3 + 1*4^3 = 2*4^3 = 2000{base 4}

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}.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q005. How would the decimal number 659 be expressed in base 4?

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Your solution:

the four place values are 4^0=1, 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024,.... since 659 falls in between 256 and

1024 we will not use any 4^5s but will use 4^4s and under. dividing 659 by 256 gives us the number of 4^4s we will

use. 659/256 = 2 with a remander of so 147. we can use only two 4^4s = 2*4^4

next we take the remainder and divide it by 64 (4^3s). 147/64 = 2 with a remainder of 19. we need two 4^3s = 2*4^3

next we divide the new remainder by 16 (4^2s). 19/16= 1 with a remainder of 3. 1*4^2

the new remainder will not divide into 4 so we are left with 0*4^1

3/1= 3 so we need 3*4^0

2*4^4 + 2*4^3 + 1*4^2 + 0*4^1 + 3*4^0 = 22103{base 4}

confidence rating #$&*:3

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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.

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q006. Find the base-10 equivalent of the number 322{base 4}.

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Your solution:

confidence rating #$&*:

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Question: `q007. Find the base-4 equivalent of the number 487.

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Your solution:

the four place values are 4^0=1, 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024,.... since 487 falls in between 4^4 and

4^5 we won't use any 4^5 but will use some 4^4s and below.

first we need to divide 487 by 256. 487/256 = 1 with a remainder of 231 so we need 1*4^4

next we divide the remainder 231 by 64. 231/64 = 3 with a remainder of 39. we need 3 4^3s. 3*4^3

next we divide 39/16= 2 with a remainder of 7. so we need 2 4^2s. 2*4^2

divide 7/4 = 1 with a remainder of 3. 1*4^1

3/1 =3 which means we need 3*4^0

1*4^4 + 3*4^3 + 2*4^2 + 1*4^1 + 3*4^0 = 13210{base 4}

confidence rating #$&*:

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Self-critique Rating:ok

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&#Good responses. Let me know if you have questions. &#