# Summarize subnet from 4 ip addresses

I am trying to figure out how to summarize the subnet for ip route. I am following this information here... http://www.pearsonitcertification.com/articles/article.aspx?p=2169746&seqNum=7

It basically is having me change the "interesting octet" into binary, and then find the common barrier between the 1 and 0. However I am unsure since one of them is 0's. The IP's are..
10.10.0.1
10.10.2.1
10.10.4.1
10.10.6.1
When I convert them into binary I get
0 = 0000 0000
2 = 0000 0010
4 = 0000 0100
6 = 0000 0110
Would it end up being 10.10.0.0/21?

Yes. The number of bits in common is the number you are looking for. You left off the first 16 bits (they are in common). plus the 5 bits in common you show here is 21 bits in common.

• Thank you. I wanted to make sure that it was the common bits and not just a wall between a 1 and 0 bit that is common in all. – Jack Hammer Sep 28 '15 at 21:40
• Start from the left, and once you hit a bit that doesn't match, you have gone one too far. – Ron Maupin Sep 28 '15 at 21:42

10.10.0000 0000.0000 0000

10.10.0000 0 are the common bits, the network bits

000.0000 0000 are the bit that are dedicated to host on your network 10.10.0.0/21

Without the binary

Where is the change happening ? - in the third octet

How big a block size do i need ? - Well its 0 - 6 in the 3rd octet, so a block size of 6 would be great but there is no such thing, so 8 will do. (block-sizes 2,4,8,16,32,64,128,256 etc etc )

Right i now know that i need a block size of 8, how many bits do i need to borrow to accomplish this? 2,4,8 - so the answer is 3 - i borrowed 3bits, 3^2 = 8

And now for some basic maths :: 24 - 3 = 21 !

Why did i subtract from 24? Because an ip address is made up of 4 * 8bits, i am working in the 3rd octet, if i was working in the 4th it would be

32 - 3 = 29 !

Anyway, 24 - 3 = 21 ! So i now know my mask is /21

My networks are ::

10.10.0.0

10.10.8.0

10.10.16.0

10.10.24.0

10.10.32.0

etc etc .