QA 17

#$&*

course Mth 151

3/6 2:24

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 5 questions in this set.

The preceding calculations have been 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 3 * 10^2 + 2 * 10^1 + 1 * 10^0

It looks like the number is 321 again.

confidence rating #$&*: 2

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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): Okay. I understand how the problem was worked out, but I don’t understand what it means when it says “base 10”. I thought that meant that I was supposed to substitute the 4s for 10s.

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Self-critique Rating:0

@&

321{base 4} = 3 * 4^2 + 2 * 4^1 + 1 * 4^0, which is equal to the base-10 number 57.

3 * 10^2 + 2 * 10^1 + 1 * 10^0 would be the base-10 number 321.

*@

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: I don’t know. I’m going to look at the answer and see if I can understand.

confidence rating #$&*: 0

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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): Okay. I’m getting it now.

@&

I figured you would, but inserted a note on that first problem anyway, just to be sure.

*@

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Self-critique Rating:0

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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: Ugh! I don’t know!

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 * 4^1 =

(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): I understand now.

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Self-critique Rating:0

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: (4 * 4^2 + 2* 4^2) + (4* 4^1 + 3 * 4^1) + (3* 4^0+ 1* 4^0)

I added the numbers 1*4^0

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

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

1* 64 + 3* 16 + 3 * 4 + 4 * 1

64 + 48 + 12 + 4

= 128

Something tells me that I did this wrong.

confidence 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}.

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Self-critique (if necessary): Yep. I got it wrong.

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Self-critique Rating:0

@&

I believe you see what you did.

*@

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 6 * 4^2 + 5 * 4^1 + 9 * 4^0

(4 * 4^2 + 2* 4^2) + (4* 4^1 + 1 * 4^1) + (4* 4^0 + 4*4^0+ 1 * 4^0)=

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

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

confidence rating #$&*: 1

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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): Shew! This stuff is giving me a fit!

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Self-critique Rating:0

@&

You expressed 659 {base 4} in base 10.

This takes a little getting used to, but you'll sort it out.

*@

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 6 * 4^2 + 5 * 4^1 + 9 * 4^0

(4 * 4^2 + 2* 4^2) + (4* 4^1 + 1 * 4^1) + (4* 4^0 + 4*4^0+ 1 * 4^0)=

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

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

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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): Shew! This stuff is giving me a fit!

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Self-critique Rating:0

@&

You expressed 659 {base 4} in base 10.

This takes a little getting used to, but you'll sort it out.

*@

#*&!

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 6 * 4^2 + 5 * 4^1 + 9 * 4^0

(4 * 4^2 + 2* 4^2) + (4* 4^1 + 1 * 4^1) + (4* 4^0 + 4*4^0+ 1 * 4^0)=

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

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

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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): Shew! This stuff is giving me a fit!

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Self-critique Rating:0

@&

You expressed 659 {base 4} in base 10.

This takes a little getting used to, but you'll sort it out.

*@

#*&!#*&!

&#Good responses. See my notes and let me know if you have questions. &#