Hack the box academy Subnet question

This question is doing my head in. I know the answer to the question but the answer fields seem to want an exact entry.

Split the network 10.200.20.0/27 into 4 subnets and submit the network address of the 3rd subnet as the answer.

What is the format that they want for the answer?

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I haven’t done the academy, but going off technical knowledge, they are probably wanting to know the network address in standard format, so 10.200.20.___. In a /27 network, you can have 8 subnets. They are wanting to know the 3rd network address from the 8 available.

Don’t worry about CIDR notation or anything, just the network address for THAT subnet.

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For this one they want it in four subnets so there will be a range of addresses and I need to know how to lay them out as in 10.200.200.___ - ___ or 10.200.20.___ - 10.200.20.___ another way I tried is 10.200.20.-. If it wasn’t for the submission seaming to need to be exact I would be done with this question

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@Macca87 said:

For this one they want it in four subnets so there will be a range of addresses and I need to know how to lay them out as in 10.200.200.___ - ___ or 10.200.20.___ - 10.200.20.___ another way I tried is 10.200.20.-. If it wasn’t for the submission seaming to need to be exact I would be done with this question

Have you tried using /29 as the mask (I haven’t done this lab either, but if I was asked to divide 10.10.10.0/27 into four, it would be 10.10.10.0/29 etc.)

That is indeed how I’m getting my answer, so I know what the answer is but it isn’t accepting it

Solved. The question only wants the first address of the third subnet not the range. That has done my head in for far longer than I am proud to say.

I’m on the same boat as you, I find out that you solved, but I’m some lost on this one lol, I really hate this questions that requires a exact entry

Type your comment> @Macca87 said:

Solved. The question only wants the first address of the third subnet not the range. That has done my head in for far longer than I am proud to say.

Hi bro, in what format the address should be input? I try xx.xxx.xx.xx, xx.xxx.xx.xx/yy, xx, xx/yy nothing works.

It’s pretty basic. Think of the range you get from a /27 network. 32 addresses, minus 2 for the network (0) and the broadcast (31), means possible ips are 10.10.10.1 to 10.10.10.30. You need to break this whole range into 4, so take the 32 total addresses and divide by 4. Now you have network address 10.10.10.0 to broadcast 10.10.10.7, then network address 10.10.10.8 to broadcast 10.10.10.15 then network 10.10.10.16 to broadcast 10.10.10.23, then network 10.10.10.24 to broadcast 10.10.10.31.

I hope this helps. :slight_smile:

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@TazWake said:
@Macca87 said:

(Quote)
Have you tried using /29 as the mask (I haven’t done this lab either, but if I was asked to divide 10.10.10.0/27 into four, it would be 10.10.10.0/29 etc.)

Breaking the /27 into 4 subnets would result in 4 separate /29 ranges. 0/29, 8/29, 16/29 and 24/29.

Might be easier to just do the math in your head for such a small range.

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Answer - 10.200.20.16

So the reason this question was giving me a hard time was because I was not using the full address. You have to split the 32 addresses into 4 equal parts, so 8 addresses per part. I was starting at 10.200.20.8 instead of starting at 10.200.20.0 because this address is still a part of the address space.

(Its says I’m responding to Androwranni, but this is just a general message)

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Hey, I used an ipv4-subnetcalculator. Answer is 10.200.20.16. Hope this works

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Did anyone get the answer for the 2nd subnet as i have tried a few different ones and not working?

@Andowrannl … told us … 10.200.20.15