# A Question?

MemberMember Posts: 57 ■■□□□□□□□□
Hi. Am new to CCNA forum.Was just wondering what the ans to this question is?

10.0.0.0/26

S=262144
Hosts per Subnet=62
Block Size=64

How do you find the 262138th Subnet directly from a formula (including the first host,the last host,the broadcast address) instead of like working backwards manually from 262144th Subnet?

The (Subnet position number-1)*block size supposedly does not stand valid ?Or am i missing something?

signing out--ritamshome.

• Senior Member Member Posts: 277
Omitting my question. I think the crack is getting to me!
"To realize one's destiny is a person's only obligation."
• Senior Member Member Posts: 135
I think working backwards would have to be your best bet.

10.0.0.0/26

Here are our bit boundries for that address space.

0-63
64-127
128-191
192-255

262144 - 262138 = 6

So we need to go back 6 subnets from 10.255.255.255 to get the one we need.

4 /26's per /24 so that puts us half way through

4 of them are in 10.255.255.0/24, and we are half way through 10.255.254.0 (at 10.255.254.12

So our subnet should be 10.255.254.128/26

10.255.254.128 - network
10.255.254.129 - first host
10.255.254.190 - last host

here is a list of bit boundries, you can't use them for the exam, but they are useful in "real life"

0-255

0-127
128-255

0-63
64-127
128-191
192-255

0-31
32-63
64-95
96-127
128-159
160-193
192-223
224-255

0-15
16-31
32-47
48-63
64-79
80-95
96-111
112-127
128-143
144-159
160-175
176-191
192-207
208-223
224-239
240-255

0-7
8-15
16-23
24-31
32-39
40-47
48-55
56-63
64-71
72-79
80-87
88-95
96-103
104-111
112-119
120-127
128-135
136-145
146-151
152-159
160-167
168-175
176-183
184-191
192-199
200-207
208-215
216-223
224-231
232-239
240-247
248-255

/30 is not listed because there are to omany, but they are easy because the network address is anything divisible by 4.

eg. if you had 10.10.10.14/30 you know that is a host address because 12 is divisible by 4 but 14 is not.
• Member Member Posts: 57 ■■□□□□□□□□
Thanks for answering. You are quite right. Working backwards seems to be the answer.-Although tedious.Heck, i am still working on the formula to directly find out the ans.Will inform all as soon as i finish working it out.

signing out--ritamshome.