Which of the following is the minimum subnet mask to allow 172.16.1.5 and 172.16.4.3 to be on the same subnet?

The correct answer is B. /21. To find the minimum (widest) subnet containing both addresses, compare the binary of the third octets: 1 = 00000001 and 4 = 00000100. They share the same first 5 bits (00000xxx), so the network prefix must be at least 16 + 5 = 21 bits. With /21 (255.255.248.0): 172.16.1.5 →…

Networking concepts

Question

Options

A/19
B/21
C/22
D/24

How the community answered

(48 responses)

A
10% (5)
B
65% (31)
C
4% (2)
D
21% (10)

Explanation

To find the minimum (widest) subnet containing both addresses, compare the binary of the third octets: 1 = 00000001 and 4 = 00000100. They share the same first 5 bits (00000xxx), so the network prefix must be at least 16 + 5 = 21 bits. With /21 (255.255.248.0): 172.16.1.5 → network 172.16.0.0, and 172.16.4.3 → network 172.16.0.0. Both fall in the same /21 subnet (172.16.0.0 – 172.16.7.255). With /22 (255.255.252.0): 172.16.1.5 → 172.16.0.0/22 and 172.16.4.3 → 172.16.4.0/22 - different subnets. Therefore /21 is the minimum mask that places both hosts on the same subnet.

Topics

#subnetting#CIDR#subnet calculation#IPv4

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