Question

FFFF will be the last memory location in a memory of size

• A. 1k
• B. 64k
• C. 32k
• D. 16k

Hint:

A hexadecimal address with 'n' digits represents \( 16^n \) total locations (from 0 to \( F...F \)). So, a 4-digit hex address (like FFFF) corresponds to \( 16^4 \) locations. Since \( 16^4 = (2^4)^4 = 2^{16} \) and \( 1K = 2^{10} \), the size is \( 2^6 \times 2^{10} = 64K \).

Answer 1

The Correct Option is B

The address `FFFF` is given in hexadecimal (base 16). To find the size of the memory, we need to determine the total number of unique addresses.
The addresses range from `0000` to `FFFF`.
The total number of addresses is `FFFF` + 1.
\[ FFFF_{16} + 1_{16} = 10000_{16} \] Now, we convert this number to base 10 to find the total number of locations.
\[ 10000_{16} = 1 \times 16^4 + 0 \times 16^3 + 0 \times 16^2 + 0 \times 16^1 + 0 \times 16^0 \] \[ = 16^4 \]
Since computer memory sizes are based on powers of 2, let's express this in base 2.
\[ 16^4 = (2^4)^4 = 2^{16} \] So there are \( 2^{16} \) total memory locations.
In computer terminology, 1K (kilobyte or kilolocations) = \( 2^{10} \).
We can rewrite \( 2^{16} \) in terms of K: \[ 2^{16} = 2^6 \times 2^{10} = 64 \times 2^{10} = 64\text{K} \] The memory size is 64K.