Homework Help, little help plz Options

[quest4547]

link

i just want to make sure im doing this stuff right, thanks in advance

[quest4547]

so far im having trouble with converting 1023 any suggestions?

[Rylinkus]

1023 is a 16 bit number

we're talking

ones place 2^0
2's place 2^1
4's place 2^2
8's place 2^3
etc

so 2^10 is 1024 so it MUST be less then

0000010000000000

2^9 would be 512
512- 1023 leaves........511
So we know the number must be

0000001XXXXXXXXX

2^8 = 256
so we have
00000011XXXXXXXX

etc.... til you get

00000011 11111111

Or you could just say

1024 = 00000100 00000000
If you subtract 1 you get.....

00000011 11111111

You could even do that out long hand with all the borrowing and shit.

[quest4547]

for number 3 the first problem where the binary number is 000011110001111 would that equal 78f in hexadecimal?


edit: and thanks alot Rylinkus for the help

This post has been edited by quest4547: Feb 14 2005, 05:21 AM

[quest4547]

anyone

[quest4547]

6. Devise a numbering system based on 5 and represent your age in that system.

any help with this problem, not sure how to go about it

[Rylinkus]

All any number system is a system where each progressive slot represents the numbers value raised to said exponent.

So a base X number system would workl as follows.....

the first place = X ^0s or the ones place with ANY base number
2nd place is X ^1s place. (We're used to 10 ^1 in base 10 systems)
3rd is X ^ 2s place (Think 10 ^2 or 100s place in our system)
And so on

Now any number is formed using these digits and their values. So the number 283 would be equal to. (2*100)+(8*10)+(3*1)

So a base 5 system the first digitis
5^0 place = 1s place
next....
5^1s place = 5s place
next.....
5^2s place = 25s place

So let's assume the digits are X,Y,Z as the number XYX

So MY age in base 10 is 25.....

So in this system the X place is 5^2 or 25. Well My age is 25 so I'd put a "1" in for that digit.
So we have 1YZ.....
Now if we subtract the 25 represented by the 1s digit from my age in base 10 we have 25-25 - 0. So we have no years left to represent and thusly the last to digits become 0.

So I'd be 100 in base 5. (God that seems old.)

Now if you were say 22..... let's start again.... XYZ
well the 25s place would have to be zero since 22 is less then zero.
So the first digit would be 0 yielding 0YZ
So we still have 22 left to represent so we move on to the second digit
Y represents the 5^1s place which simplifies to the 5s place.
So we can get four 5's in 22.
This makes the 2nd digit 4. 04Z
4*5 = 20 which will leave 2 to represent.....
Well Z is the ones place and we have [2] 1's so we need to make Z = 1.

This gives us the number 0 4 2. Now if we break this down using what the digits mean we can check ourselves. by converting to base 10.....

(0 * (5 ^ 2)) + (4 * (5 ^ 1)) + (2 (5 ^ 0))
So Ill do the work in the inner most parentheses and get rid of the exponentials and get
(0*25)+(4*5)+(2*1)
So we'll do the multiplication next....
0+20+2 = 22 (which is good since its what we were trying to represent)

Hope that helps.
If you ARE 22 or 25 you're golden

[BinaryMod]

whoo hoo go salukis.... lol

[quest4547]

whoo hoo go salukis.... lol

hell yea.....lol

thanks again Rylinkus

[Rylinkus]

no prob dude. Only hope Im helping.