GATE EC 2026 Set 1 — Question 42

Given:
First address of the memory = $(2500)_H$
Memory size = 256 KBFirst, convert the memory size from KB to bytes:
1 KB = $2^{10}$ bytes = 1024 bytes
256 KB = $256 \times 1024$ bytes = $2^8 \times 2^{10}$ bytes = $2^{18}$ bytesTo find the number of addresses, we need to express $2^{18}$ in hexadecimal. Since $2^{18} = 2^2 \times 2^{16} = 4 \times (2^4)^4 = 4 \times (16)^4$, this means the memory occupies $40000_H$ addresses.Alternatively, $2^{18}$ bytes is $2^{18}$ unique addresses. In hexadecimal, each digit represents 4 bits. So, $2^{18}$ bytes requires 18 bits for addressing. This means the addresses will range from $0$ to $2^{18}-1$. The number of locations is $2^{18}$.Let's convert $2^{18}$ to hexadecimal:
$2^{18} = 262144$ in decimal.
To convert to hexadecimal, divide by 16 repeatedly or group bits.
$2^{18} = (1000000000000000000)_2$
Group into 4 bits from the right:
$0010 \ 0000 \ 0000 \ 0000 \ 0000_2$
This is $2 \ 0 \ 0 \ 0 \ 0_H$.
So, 256 KB corresponds to $40000_H$ locations (from $00000_H$ to $3FFFF_H$).The number of memory locations is $2^{18}$. If the first address is $A_{first}$, and there are $N$ locations, the last address is $A_{first} + N - 1$.
Here, $N = 2^{18}$ bytes. So, the last address will be $A_{first} + (2^{18} - 1)$ bytes.First address = $(2500)_H$
Number of locations = $(40000)_H$Last address = First address + (Number of locations - 1)
Last address = $(2500)_H + (40000_H - 1_H)$
Last address = $(2500)_H + (3FFFF)_H$Now, perform hexadecimal addition:
$2500_H$
+ $3FFFF_H$
----------
$424FF_H$Let's verify:
$0 + F = F$
$0 + F = F$
$5 + F = 14_H$ (decimal 20) = $4_H$ with carry $1_H$
$2 + 3 + 1 (carry) = 6_H$So, the last address is $(424FF)_H$.The final answer is $\boxed{\text{C}}$