Convert the octal number 73 (base 8) to its binary (base 2) equivalent by expanding each octal digit into a 3-bit group.

Introduction / Context:
Octal-to-binary conversion is straightforward because each octal digit corresponds to exactly three binary bits. This mapping is widely used in low-level work, especially where groups of three bits align nicely with permissions, opcodes, or compact numeric notation.

Given Data / Assumptions:

• Octal value: 73 (base 8).
• Digit 7 and digit 3 must be mapped to 3-bit binary groups.
• No leading groups are omitted.

Concept / Approach:
Map each octal digit separately: 0→000, 1→001, 2→010, 3→011, 4→100, 5→101, 6→110, 7→111. Concatenate the resulting 3-bit groups in the same order to produce the binary value.

Step-by-Step Solution:
7 (octal) → 111 (binary).3 (octal) → 011 (binary).Concatenate: 111 and 011 → 111011.Select “111011.”

Verification / Alternative check:
Decimal cross-check: 73 (octal) = 7*8 + 3 = 56 + 3 = 59. Now binary 111011 = 32 + 16 + 8 + 2 + 1 = 59. Both methods agree.

Why Other Options Are Wrong:
110111 and 111100 represent different decimal values; 110010 is 50 decimal; “None” is wrong because 111011 is correct.

Common Pitfalls:
Dropping a leading 0 in the lower 3-bit group (e.g., writing 3 as 11 instead of 011) and thus producing a 5-bit result; reversing the order of digits.

Final Answer:
111011