Question

Find the number of addresses in a range if the first address is 146.102.29.0 and last address is 146.102.32.255.

(a) 1028

(b) 1024

(c) 578

(d) 512

I had been asked this question in an international level competition.

This is a very interesting question from IP Addressing in section Internet Protocol, Network Addressing and Transport Level Security of Cryptograph & Network Security

Answer 1

Correct option is (b) 1024

Explanation: Subtract the first address from the last address in base 256.

The result = 0.0.3.255

Therefore, number of addresses = (3 X 256 + 255) + 1 = 1024.