Binary to Octal — Group Bits in Threes Find the octal equivalent of the binary number 1101110101110110 by grouping into 3-bit sets from the right.

Introduction / Context:
Converting binary to octal is efficient because 8 equals 2^3. Therefore, every three binary bits map to exactly one octal digit. This technique avoids full decimal conversion and reduces working time and errors.

Given Data / Assumptions:

• Binary input: 1101110101110110.
• No sign or fractional portion.
• Left padding with zeros is permissible to complete a leading triplet if required.

Concept / Approach:
Partition the binary string into 3-bit groups starting at the least significant end. Translate each triplet to an octal digit using 000→0 through 111→7 and assemble the digits from most significant group to least significant group.

Step-by-Step Solution:
1) From the right, group bits: 1 101 110 101 110 110 → add leading zeros to the lone 1 to form 001.2) Final grouping: 001 101 110 101 110 110.3) Translate: 001→1, 101→5, 110→6, 101→5, 110→6, 110→6.4) Combine: 156566 base 8, written as 1565668.

Verification / Alternative check:
Convert 1565668 back to binary: 1→001, 5→101, 6→110, 5→101, 6→110, 6→110. Concatenate to get 001101110101110110 and remove the leading zeros to return to 1101110101110110.

Why Other Options Are Wrong:

• 6545218, 5565618, 1566568: These imply different triplet translations or wrong group alignment and do not reproduce the given binary when mapped back.

Common Pitfalls:
Grouping from the left side, skipping the leading zero padding for an incomplete initial group, or reversing the order of triplets when writing the octal digits.

Final Answer:
1565668