The amount of cells in a computer`s main memory

[guitar]

Homework Statement

How many cells can be in a computer`s main memory if each cell`s address can be represented by 2 hexadecimal digits? what if 4 hexadecimal digits are used?

Homework Equations

N/A

The Attempt at a Solution

ok, i tried and other people helped and i am stuck with this:

1. If each cell's address is only 2 hex digits, you are using 16 binary digits, and therefore 65,536 bytes.

2. Hexadecimal means base 16. In other words, 16 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B ,C, D, E, F).
Two hexadecimal digits give you 16 x 16 = 256 locations.
Four hexadecimal digits give you 16 x 16 x 16 x 16 = 65536 locations.

3. 0 (hexadecimal representation) is 0000 (bit pattern)
1 (hexadecimal representation) is 0001 (bit pattern)
2 (hexadecimal representation) is 0010 (bit pattern)
etc etc etc

i am trying to connect the dots here...are those 16 digits in base 16- 16 locations? also are these digits: (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B ,C, D, E, F) 4 bits or 8 bits each? or are they just 1 bit each? and how does the hex to bit pattern representation play up into this?

thanks.

 

[Mark44]

i am trying to connect the dots here...are those 16 digits in base 16- 16 locations? also are these digits: (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B ,C, D, E, F) 4 bits or 8 bits each? or are they just 1 bit each? and how does the hex to bit pattern representation play up into this?

I don't understand what you're asking in your first question. What do you mean "are those 16 digits in base 16- 16 locations?"
Each hex digit can be represented by four bits:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
...
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111

Any hex digit 8 or higher requires 4 bits to represent it. Smaller hex digits could be represented with fewer bits. The only numbers that could be represented with one bit are 0 and 1.

 

[guitar]

I don't understand what you're asking in your first question. What do you mean "are those 16 digits in base 16- 16 locations?"
Each hex digit can be represented by four bits:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
...
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111

Any hex digit 8 or higher requires 4 bits to represent it. Smaller hex digits could be represented with fewer bits. The only numbers that could be represented with one bit are 0 and 1.

for starters i see 4 digits for all the hex digits from 0 to F. for example bit pattern for the hex digit 3 is 0011(4 bits)...please elaborate on this.

 

[turin]

How many cells can be in a computer`s main memory if each cell`s address can be represented by 2 hexadecimal digits? what if 4 hexadecimal digits are used?
...
1. If each cell's address is only 2 hex digits, you are using 16 binary digits, ...

I don't follow this. Each hex digit specifies four binary digits, so two hex digits specifies (two times four equals) eight binary digits.

2. Hexadecimal means base 16. In other words, 16 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B ,C, D, E, F).

Yes, there are sixteen possible values for each hex digit.

Two hexadecimal digits give you 16 x 16 = 256 locations.
Four hexadecimal digits give you 16 x 16 x 16 x 16 = 65536 locations.

Doesn't this already answer the original problem?

are those 16 digits in base 16- 16 locations?

I suppose that the sixteen possible values of the least significant digit (LSD :O) distinguish sixteen memory locations, given some fixed value for any other more significant digits. Each additional hex digit in the number multiplies the number of possible distinct values by sixteen. Your wording here is very confusing, though - too many "16"s.

also are these digits: (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B ,C, D, E, F) 4 bits or 8 bits each?

You only need up to four bits to specify one of these hex digits. Each hex digit requires four bits.

how does the hex to bit pattern representation play up into this?

I don't know what you're talking about.