Assignment 20

course Mth 151

nGX֧Ӱassignment #020

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Liberal Arts Mathematics I

07-16-2006

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15:20:39

query 4.3.6 number following base-six 555

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RESPONSE -->

557 base six

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15:22:55

** 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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RESPONSE -->

Oh. I got that one all wrong.

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15:23:49

query 4.3.20 34432 base five

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RESPONSE -->

3*5+4*5+4*5+3*5+2 = 2492

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15:23:57

**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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RESPONSE -->

ok

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15:25:38

Explain how you use the calculator shortcut to get the given number.

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RESPONSE -->

You multiply each number by the base (in the last example, 5) and then add the next number to be multiplied by the base. This way you can keep on using the calculator instead of having to write anything down. Don't multiply the very last number by the base.

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15:26:03

** 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 5s 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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RESPONSE -->

ok

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15:33:03

query 4.3.40 11028 decimal to base 4

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RESPONSE -->

11028

2757 R0

689 R1

172 R0

43 R3

10 R2

2 R2

0 R2

All of the remainders are used to make the number 2223010.

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15:35:58

** 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.

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RESPONSE -->

I thought that we had to divide for this problem. Can you do this problem both ways? I came close to the correct answer, but I messed up somewhere.

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15:37:11

0 * 4^1.

Thus our number is 2230110 base 4.

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RESPONSE -->

I'm not quite sure where I messed up, but I got close to the answer, so at least I'm not out in left field.

I would have to redo your arithmetic to find the error, but I see nothing wrong with your method.

I do confirm, however, that 2230110 is the correct result.

Actually I do see your error. When you divide 0 by 4 you get 0 with remainder 0. You don't get remainder 2 on that last step.

It's best to understand the solution the way I give it. The sequential-division method of changing bases works, but it's harder to understand why it works and therefore harder to see when you've made an error.

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15:41:58

query 4.3.51 DC in base 16 to binary

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RESPONSE -->

2441 - 0

1220 - 1

610 - 0

305 - 0

152 - 1

76 - 0

38 - 0

19 - 1

9 - 1

4 - 0

2 - 0

1 - 1

100110010010

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15:43:51

** 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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RESPONSE -->

Golly, I thought that I understood the chapter when I read it, but I kept messing up on these questions. I understand how to get the number, but I must be forgetting how to do it on my own.

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15:45:45

Is 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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RESPONSE -->

Sometimes even and sometimes odd, because the first numbers are 1, 2, 3, ... 8.

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15:47:23

** 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. **

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RESPONSE -->

ok

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"

You did well on this assignment. See my notes and let me know if you have questions.