312-50V9 · Question #516
An nmap command that includes the host specification of 202.176.56-57.* will scan _______ number of hosts.
The correct answer is C. 512. The nmap range 202.176.56-57.* specifies 2 values for the third octet and 256 values (0-255) for the fourth octet, yielding 2 x 256 = 512 total hosts.
Question
An nmap command that includes the host specification of 202.176.56-57.* will scan _______ number of hosts.
Options
- A2
- B256
- C512
- DOver 10, 000
How the community answered
(28 responses)- A4% (1)
- C93% (26)
- D4% (1)
Why each option
The nmap range 202.176.56-57.* specifies 2 values for the third octet and 256 values (0-255) for the fourth octet, yielding 2 x 256 = 512 total hosts.
2 would only be correct if the wildcard '*' resolved to a single value, but '*' expands to all 256 possible values (0-255) for that octet.
256 would result if only one value were specified for the third octet with a wildcard fourth octet, but the range '56-57' contributes a factor of 2, doubling the count to 512.
In nmap's host specification syntax, '56-57' means the third octet takes exactly 2 values (56 and 57), while '*' is shorthand for the full range 0-255, giving 256 values for the fourth octet. Multiplying these two ranges together produces 2 x 256 = 512 unique IP addresses. This is standard nmap notation for specifying arbitrary IP ranges without using CIDR notation.
Over 10,000 hosts would require at least two fully wildcarded octets; only one octet is wildcarded here, capping the total at 512.
Concept tested: Nmap host range specification and IP address counting
Source: https://nmap.org/book/man-target-specification.html
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