Question

How many address bits are required to represent a 32k *12 memory?

Answer 1

You need log2(n) bits to address n bytes. For example, you can store 256 different values in an 8 bit number, so 8 bits can address 256 bytes. 2e10 = 1024, so you need 10 bits to address every byte in a kilobyte. Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 2e32 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory.