GATE CS 2026 Set 1 Q.56: IPv4/IPv6 Addressing, Subnetting, CIDR, ARP, ICMP and NAT

An ISP having an address block 202.16.0.0/15 assigns a block of $6000$ IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?

202.16.0.0/19

202.17.64.0/19

202.16.32.0/19

202.17.24.0/19

Official answer and solution

Official answer: A, B and C
From: Official GATE 2026 answer key

Solution

Check both inclusion within 202.16.0.0/15 and CIDR prefix alignment (the network address of a /19 block must be a multiple of $2^{32-19} = 2^{13}$ ).

Prefix length: To allocate at least $6000$ IP addresses, the block size must be $2^{13} = 8192 \ge 6000$ , which corresponds to a /19 prefix ( $32 - 13 = 19$ ).
Range of ISP block 202.16.0.0/15: Covers 202.16.0.0 through 202.17.255.255.
Alignment requirement for /19: In the third octet (bits 17 to 24), the first $19 - 16 = 3$ $19 - 16 = 3$ bits belong to the network prefix and the remaining $5$ $5$ bits belong to the host part and must be $0$ $0$ . Therefore, the third octet of a valid /19 block address must be an integer multiple of $2^5 = 32$ $2^{5} = 32$ ( $0, 32, 64, 96, 128, 160, 192, 224$ $0, 32, 64, 96, 128, 160, 192, 224$ ):
- (A) 202.16.0.0/19: Third octet is $0$ (multiple of $32$ ) $\rightarrow$ Valid.
- (B) 202.17.64.0/19: Third octet is $64$ (multiple of $32$ ) $\rightarrow$ Valid.
- (C) 202.16.32.0/19: Third octet is $32$ (multiple of $32$ ) $\rightarrow$ Valid.
- (D) 202.17.24.0/19: Third octet is $24$ (not divisible by $32$ ) $\rightarrow$ Invalid.

Final answer

Options (A), (B), (C)

Common trap

Selecting (D) by checking only that 202.17.24.0 lies inside 202.16.0.0/15 without checking /19 boundary alignment.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question: 56 of 65
Section: Computer Science and Information Technology

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