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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.

Scanning Networks

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)
  • A
    4% (1)
  • C
    93% (26)
  • D
    4% (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.

A2

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.

B256

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.

C512Correct

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.

DOver 10, 000

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

Topics

#nmap host specification#IP range syntax#host count calculation#wildcard notation

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