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17 address lines and 8 data lines.

2^17=128k

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Q: How many address lines and data lines are required for a 128k x 8 memory system?
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How many address lines are needed to access 256KB of main memory?

The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.


How many address lines and data lines are required for a2Kx8 memory?

A 2K X 8 memory requires 11 address lines and 8 data lines


How many no of address lines required in 1MB memory 111622 or 24?

How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.


How many nuMBer of address lines required for 8 MB of memory?

It takes 23 address lines to address 8 mb of memory.


How many address lines are required if micro processor has to access 2kb memory?

2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required


How many address lines are required to address 64 kbytes of memory?

You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.


How many address lines are necessary to address two megabytes of memory?

Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11


How many address lines can access a 16MB memory?

for 16 MB memory has 24 address lines


How many address lines would be required for a 2K?

In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..


How many nuMBer of address lines required for 1 MB memory?

Firstly we need to convert Mb's into bits i.e 1Mb=1024x1024 = 210x210 =220 That means there are 220 memory locations and we will need 20 address lines.


How many address lines are require to access 128MB of memory?

ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.


How many no of address lines are required for 512 K word memory?

If you are addressing bytes, then 512K words (16 bit words) requires 20 address lines.I gave that answer because the question was categorized 8086/8088. If you are addressing words, then the answer is 19 address lines.