GPEN Question #198: Real Exam Question with Answer & Explanation

The correct answer is C. 512. The range 56-57 in the third octet yields 2 values and the wildcard in the fourth octet yields 256 values, totaling 512 hosts.

Question

You want to run the nmap command that includes the host specification of 202.176.56-57.*. How many hosts will you scan?

Options

A1024
B256
C512
D64

Explanation

The range 56-57 in the third octet yields 2 values and the wildcard in the fourth octet yields 256 values, totaling 512 hosts.

Common mistakes.

A. 1024 would require four values across the specified octet range multiplied by 256, which does not match the 56-57 range.
B. 256 would be correct only if a single value were specified in the third octet rather than a two-value range.
D. 64 does not correspond to any valid arithmetic result from the given host specification.

Concept tested. Nmap host range specification and scan count calculation

Reference. https://nmap.org/book/man-target-specification.html

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