# IP Subnetting help

Can this IP address, 10.3.104.95 exist on this subnet  10.3.100.0/22? Thank you.

## Comments

• A /22 will give you the following bit pattern in the 3rd octect: 1111 1100. This translates to 128+64+32+16+8+4 which gives 252. So the mask is 255.255.252.0. You then subtract 252 from 256. This gives you the subnet id increment of 4 (the cisco book call this the magic number).

100 can be divided by 4 without remainder so you know 10.3.100.0 is a valid subnet id given the subnet mask. You know the increment is 4.

Can you complete?

• Can this IP address, 10.3.104.95 exist on this subnet  10.3.100.0/22? Thank you.

Hello,

10.3.104.95 will not belong to 10.3.100.0/22.

EDIT: My original wording was bad. Sorry for the confusing way of describing things originally. There are 1024 addresses in 10.3.100.0/22. The last address in that subnet will be the broadcast address 10.3.103.255. Since the value in your 3rd octet is 104 and that's bigger than the 3rd octect for the last address in this network, you know that it can not belong to 10.3.100.0/22.

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The network is 10.3.100.0/22.

First Host Address is 10.3.100.1

Number of Hosts: 1022 (2 to the 10th - 2) = 1022

Last Possible Host is  10.3.103.254

Broadcast Address  is 10.3.103.255

Answer still is that 10.3.104.95 would not fall within that network

• Thank you.

• You can think of a /22 as 4x /24’s— hence that prefix would cover the following range:
10.3.100.0 - 10.3.103.255

I have found http://subnettingpractice.com/ to be useful for testing my subnetting theory with practical challenges, give it a shot

- Michael

On 22 September 2016 at 10:17:28 AM, ak47inusa (bounce-ak47inusa@ieoc.com) wrote:

Can this IP address, 10.3.104.95 exist on this
subnet  10.3.100.0/22? Thank
you.

INE - The Industry Leader in CCIE Preparation

http://www.INE.com

• ✭✭✭

Lots of vidoes on u-tube that do show subnetting (learn how to) and practice site for subnetting (with question and answers)

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