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

Answer:

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

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.