UNIX permission bits (octal): For a mode of 652 (owner=6, group=5, others=2), which statement is true about the resulting permissions?

Introduction / Context:UNIX file permissions are often represented in octal form as owner-group-others. Each digit sums the bits: read=4, write=2, execute=1. Correctly interpreting these digits ensures accurate security settings and prevents accidental exposure or loss of access.

Given Data / Assumptions:

• Octal mode is 652.
• Owner digit 6 means read + write.
• Group digit 5 means read + execute.
• Others digit 2 means write only.

Concept / Approach:Break 652 into three classes. Owner=6 maps to r and w (4+2). Group=5 maps to r and x (4+1). Others=2 maps to w only. Validate each provided statement against these derived permissions to find the single correct description.

Step-by-Step Solution:

Verification / Alternative check:Use a quick mnemonic: 7=rwx, 6=rw-, 5=r-x, 4=r--, 2=-w-, 1=--x. This cross-check confirms that others have write permission in 652.

Why Other Options Are Wrong:Execute permission for the owner: owner has rw-, no execute. Read and write permission of groups: group has r-x, not write. All of the above/None of the above: inconsistent with the analysis; only the statement about “write for others” is true.

Common Pitfalls:Confusing group and others digits; assuming “5” includes write; forgetting that octal 2 is write-only. Always decompose each digit into 4, 2, and 1 to avoid mistakes.

Final Answer:Write permission for others