Exam SY0-501 topic 1 question 172 discussion

because 192.168.1.50 is not a network address in this case, you need the network address

Hey. This is my dumb head way. /24 is a 255.255.255.0 subnet so thats 256 addresses. If you make the / smaller the subnet gets more addresses and the mask gets smaller. Use this as your base, and just adjust from /24. Also remember the available addresses is always 2 lower than the total addresses. a /23 would double the addresses to 512 and the subnet mask would become 255.255.254.0 a /22 would double the addresses available again to 1024 and the mask would get 1 smaller, 255.255.253.0. On the flip side a /25 is going to halve the addresses from 256 to 128. Figuring out the subnet for /## greather than /24 you instead take 256-[addresses]. So a /25 would be 255.255.255.[256-128) A /26 would be 256/2/2 so 64 addresses available, and the mask would be 255.255.255.[256-64). You can always look up tables, or figure out the binary math... but for doing it in my head i just revert back to a /24 and adjust.

Actually, this is what helped me more so than anythign ive seen. 1. Subtract number after the slash by 32 (ex. 32-28=4) 2. Now take 2 raised to the power of that answer (2^4 =16) 3. Now subtract that answer from 2 (16-2=14) I have no prior BG in IT whatsoever and I learned this recently