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How to find a Usable IP address with just a Default Gateway and Subnet mask?

I'm working on learning how to manually solve some issues related to subnetting.
A practice question I came across states:

If you have a router in the network you're connected to and the Default Gateway is set to 192.168.1.1 and the subnet mask is 255.255.255.0, which of the following are usable IP address on that network?

1) 192.168.1.1
2) 192.168.1.300
3) 192.168.1.30
4) 192.168.2.10

Just looking at it, I can tell it's not 1), because that'd be a conflict, right?
2) would be too high of a 4th octet
4) is not the right subnet?

What would be the formula to workout the answer? Or is there not one?

I do not know what the MagicNumber would be in this case.
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garryshape
Asked:
garryshape
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3 Solutions
 
Dirk KotteSECommented:
your answer is ok and your explanation correct. (address 3 is possible)
wit a subnet mask of 24 bit 8255.255.255.0) you are free to use the nearly the full last octet (0-255) for clients.
the x.x.x.0 is the network-address and should not be used, the x.x.x.255 is the broadcast address. (used for ARP-broadcasts for example)
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davorinCommented:
You are right, the only usable IP address is 3) 192.168.1.30.

Subnet mast 255.255.255.0 tells you that you have a network with 255 IP addresses. Only the last number is changing - from 192.168.1.0 to 192.168.1.255.
You can exclude from usable adresses:
192.168.1.0 - which is the network address
192.168.1.255 - which is broadcast address
and 192.168.1.1. - which is already used by gateway.
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Dirk KotteSECommented:
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garryshapeAuthor Commented:
Thanks guys I thought maybe it was a trick question, but these explanations make more sense to me.
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