If we want to be able to add any four 16 bit unsigned numbers and store the result in a register, what should be the minimum width of the register to hold the largest result you might encounter.

Question

If we want to be able to add any four 16 bit unsigned numbers and store the result in a register, what should be the minimum width of the register to hold the largest result you might encounter.

BlackTom AI · Accepted Answer

We’re asked for the minimum register width to hold the sum of any four unsigned 16‑bit numbers.
First, note that a 16‑bit unsigned number can range from 0 to 2^16 − 1 = 65,535.
The largest possible sum of four such numbers is 4 × 65,535 = 262,140.
To store 262,140, the register must have enough bits to represent at least this value.
If a register has n bits, it can store values from 0 to 2^n − 1. We need 2^n − 1 ≥ 262,140.
Testing n = 17: 2^17 − 1 = 131,071, which is too small.
Testing n = 18: 2^18 − 1 = 262,143, which is just large enough to cover 262,140.
Therefore, the smallest width that can hold the largest possible sum is 18 bits, since 2^18 = 262,144 > 262,140 and 2^17 = 131,072 < 262,140.
This aligns with the provided option indicating an 18‑bit width.

ELEC 220 001 Exam 01

If we want to be able to add any four 16 bit unsigned numbers and store the result in a register, what should be the minimum width of the register to hold the largest result you might encounter.

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