Question

A computer’s memory is composed of 8K words of 32 bits each. How many bits are required for memory addressing if the smallest addressable memory unit is a word?

13

8

10

6

32

Answer 1

The minimum number of bits necessary to address 8K words is 13.

Step-by-step explanation:

You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.

8000 words can be addressed with
`2^(n)` units. Now you have to find the value of n that approximates to the number of words.

`2^2=4\\2^4=16\\2^7=128\\2^(10)=1024\\2^(12)=4096\\2^(13)=8192`

So you can see that 13 bits are needed to address 8K words.