# 1 More Computer Question

Discussion in 'OT Technology' started by doggfather, Sep 15, 2005.

1. ### doggfatherNew Member

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14. If you have a 100GB Harddrive, how many bits are required to address each 512-byte chunk of data?

2. ### EuclidNew Member

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I'm not sure about these numbers, i.e. maybe there are 1024 MB in a GB? But this is what I would do...

512 bytes is half of a KB, 1000 KB in a MB, 1000 MB in a GB.

So that's 2*1000*1000*100 = 200 000 000 data chunks. If each data chunk has a unique address, that's 200 000 000 addresses. How many bits do you need to count up to 200 000 000?

3. ### Slid.I'm a guy.

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We shouldn't be doing your homework for you.

4. ### EagerZeroedThickNew Member

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why would this matter in life?

5. ### spazbabyany more brain busters

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so true

6. ### ProteusOT Supporter

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Technically there are 1024 KB in a MB and 1024 MB in a GB and 1024 GB in a TB.

But yeah, who gives a fuck

7. ### peerkNew Member

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100GB = 107374182400 bytes (but HD makers usually don't use 2^30 for giga)
107374182400 bytes / 512 bytes = 209715200
2^28= 268435456

You should really do your own homework.

Last edited: Sep 16, 2005

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