How many bits does it take to store the result of two unsigned 16-bit numbers added together?Select the minimum value which will always be valid:

Break Down the Problem

1. Understanding the Bit Representation: When two unsigned 16-bit numbers are added, we need to determine the number of bits required to hold the maximum possible result.
2. Calculating the Maximum Value: The maximum value for an unsigned 16-bit number is $2^{16} - 1$.

Relevant Concepts

3. Addition of Two Numbers: When you add two unsigned integers, the maximum possible value can potentially exceed the maximum representable value of a single integer.
4. Maximum Result Calculation: If both numbers are at their maximum, the calculation is: $\text{Max Result} = (2^{16} - 1) + (2^{16} - 1) = 2^{17} - 2$

Analysis and Detail

5. Bit Requirement Calculation: To determine how many bits are needed to store this result:
   • We need the smallest $n$ such that $2^n > 2^{17} - 2$.
   • Since $2^{17} - 2 < 2^{17}$, it indicates that 17 bits are sufficient.

Verify and Summarize

6. Final Verification: A 17-bit number can represent values from 0 to $2^{17} - 1$, which is sufficient to accommodate the maximum possible result of the addition of two unsigned 16-bit numbers.

Final Answer

It takes 17 bits to store the result of adding two unsigned 16-bit numbers.