In Binary to Octal Number System Conversion, what is the octal equivalent for binary number 1101010?

Question

Accepted Answer

The correct option is (3):

The correct method for converting a binary number to its octal equivalent relies on the fact that the octal number system (base 8) is a power of two ($2^3$). This mathematical relationship means that every three binary digits can be directly translated into a single octal digit. To perform the conversion for the binary number $\mathbf{1101010}$, you must first group the bits into sets of three, starting from the rightmost digit. The initial grouping is $1 \ 101 \ 010$. Since the leftmost group has only one bit, we pad it with two leading zeros for a complete 3-bit set: $\mathbf{001 \ 101 \ 010}$. Finally, we convert each 3-bit group to its decimal/octal equivalent: $001_2 = 1_8$, $101_2 = 5_8$, and $010_2 = 2_8$. Concatenating these results yields the correct octal number, $\mathbf{152}$.

The incorrect option is (1):
The option $\mathbf{125}$ is incorrect because it corresponds to a different binary number. If we convert $\mathbf{(125)\_8}$ back to binary by replacing each octal digit with its 3-bit binary equivalent, we get $1\_8 = 001\_2$, $2\_8 = 010\_2$, and $5\_8 = 101\_2$. Combining these gives the binary number $\mathbf{001010101}$, or $(1010101)\_2$. This is not the original binary number $\mathbf{(1101010)\_2}$, which confirms that $125$ is a result of a significant error in the conversion process, likely due to incorrect grouping or a calculation mistake.

The incorrect option is (2):
The option $\mathbf{137}$ is incorrect as it represents the octal equivalent of $(1011111)\_2$. A simple back-conversion reveals this: $1\_8 = 001\_2$, $3\_8 = 011\_2$, and $7\_8 = 111\_2$. Concatenating these yields $\mathbf{001011111}$. The structure of this binary number is fundamentally different from the given $\mathbf{1101010}$, which has a sequence of two zeros in the middle ($010$). This suggests an incorrect binary-to-octal translation in the second or third group of the original number.

The incorrect option is (4):
The option $\mathbf{155}$ is incorrect and is the octal equivalent of a binary number with more ones. Converting $\mathbf{(155)\_8}$ back to binary results in $1\_8 = 001\_2$, $5\_8 = 101\_2$, and $5\_8 = 101\_2$, which gives the binary number $\mathbf{001101101}$, or $(1101101)\_2$. This is a common error where the student might have miscalculated the last binary group $\mathbf{(010)\_2}$ as $\mathbf{(101)\_2}$ (which is $5\_8$) instead of the correct $\mathbf{2\_8}$. The error lies in incorrectly evaluating the positional weight of the bits in the final triplet.