Binary to octal — group bits by threes Convert the binary number 010111100 (base 2) into its octal (base 8) representation.

Introduction / Context:
Octal and binary interconvert by grouping binary digits into sets of three, starting from the right. Each 3-bit group corresponds to one octal digit (0–7). This method is fast and avoids arithmetic division when the binary string is already available.

Given Data / Assumptions:

• Binary input: 010111100.
• We may pad the most significant side with zeros to form complete 3-bit groups.
• Octal digits map from 000→0 up to 111→7.

Concept / Approach:

Partition the binary string into 3-bit chunks from right to left, then translate each chunk to its octal digit. Keep the left-to-right order of the resulting octal digits.

Step-by-Step Solution:

Verification / Alternative check:

Round-trip: 2→010, 7→111, 4→100; concatenation recreates 010111100. The mapping is consistent.

Why Other Options Are Wrong:

1728 and 1748: first digit 1 would require leading bits 001 (not present here).

2728: middle digit 7 is correct, but the final digit 2 would require last group 010, whereas we have 100 → 4.

Common Pitfalls:

Grouping from the left instead of the right, or forgetting to pad on the leftmost side. Always ensure the total bit count is a multiple of three before conversion.

Final Answer:

2748