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What is the smallest number of bits that would be needed to store the number '50' in binary?​

Sagot :

Answer:

6 bits

Explanation:

5 bits gets you 2^5=32 possibilities (values 0-31), which is not enough for 50.

6 bits gets you 2^6=64 possibilities (values 0-63), which is enough.

To calculate for an arbitrary number, take the 2-log, and round up:

[tex]\log_2(50) \approx 5.64[/tex]

rounded up is 6.