GATE CS 1999 Q2.16 — Computer Arithmetic

To add two $16$-bit binary numbers, $A = A_{15}A_{14}\dots A_0$ and $B = B_{15}B_{14}\dots B_0$:

• Least Significant Bit (LSB) position ($0^{\text{th}}$ bit): There is no incoming carry from a previous stage. We only need to add two bits, $A_0$ and $B_0$. Therefore, $1$ half-adder is sufficient.
• Remaining $15$ bit positions ($1^{\text{st}}$ to $15^{\text{th}}$ bit): For each of these positions, we must add three bits: $A_i$, $B_i$, and the carry-in $C_i$ generated from the previous stage. Therefore, we require $15$ full-adders.

Thus, the total requirement is $1$ half-adder and $15$ full-adders.