Question:medium
The address of the first location of a 256 kilo byte (KB) memory is \((2500)_H\).
Choose the correct address of the last location of the memory.
The address of the first location of a 256 kilo byte (KB) memory is \((2500)_H\).
Choose the correct address of the last location of the memory.
Choose the correct address of the last location of the memory.
Show Hint
Last address = first address + total bytes - 1, all worked out in hexadecimal.
Updated On: Jul 20, 2026
- \((2FFF)_H\)
- \((124FF)_H\)
- \((424FF)_H\)
- \((324FF)_H\)
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Convert the starting address to decimal.
$(2500)_H=2\times16^3+5\times16^2+0\times16+0=8192+1280=9472$ in decimal.
Step 2: Find the memory size in decimal.
$256$ KB $=256\times1024=262144$ bytes.
Step 3: Find the last address in decimal.
\[ 9472+262144-1=271615 \]
Step 4: Convert 271615 back to hexadecimal.
Dividing repeatedly by powers of $16$: $271615=4\times65536+9471$, then $9471=2\times4096+1279$, then $1279=4\times256+255$, and $255=15\times16+15$. Reading the digits gives $4,2,4,15,15$, that is, $(424FF)_H$ since $15$ in hex is $F$.
Step 5: Conclude.
\[ \boxed{(424FF)_H} \] This confirms the same answer reached through pure hex addition.
$(2500)_H=2\times16^3+5\times16^2+0\times16+0=8192+1280=9472$ in decimal.
Step 2: Find the memory size in decimal.
$256$ KB $=256\times1024=262144$ bytes.
Step 3: Find the last address in decimal.
\[ 9472+262144-1=271615 \]
Step 4: Convert 271615 back to hexadecimal.
Dividing repeatedly by powers of $16$: $271615=4\times65536+9471$, then $9471=2\times4096+1279$, then $1279=4\times256+255$, and $255=15\times16+15$. Reading the digits gives $4,2,4,15,15$, that is, $(424FF)_H$ since $15$ in hex is $F$.
Step 5: Conclude.
\[ \boxed{(424FF)_H} \] This confirms the same answer reached through pure hex addition.
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