UGC NET Questions (Paper – 1)

The decimal 4095 is 2¹² − 1, which is also one less than 8⁴, so in octal it is represented by four 7s. 7777₈ expands to 7×8³ + 7×8² + 7×8¹ + 7×8⁰. This equals 3584 + 448 + 56 + 7 = 4095. Thus 7777 is the correct octal representation of 4095.

Option A:
7770₈ equals 7×512 + 7×64 + 7×8 + 0 = 3584 + 448 + 56 + 0 = 4088. Since this value is slightly less than 4095, it cannot represent the given decimal. It is a close distractor but not exact.

Option B:
7707₈ converts to 7×512 + 7×64 + 0×8 + 7 = 3584 + 448 + 0 + 7 = 4039. This value is again less than 4095, making it an incorrect conversion. The zero in the 8¹ place reduces the total.

Option C:
7077₈ stands for 7×512 + 0×64 + 7×8 + 7 = 3584 + 0 + 56 + 7 = 3647. This is much smaller than 4095, so it cannot be correct. Missing 7 in the 8² position significantly decreases the value.

Option D:
7777₈, when expanded, yields 3584 + 448 + 56 + 7 = 4095. Because this matches the specified decimal exactly, it is the correct answer. It also illustrates that a sequence of maximal digits represents the largest value for a fixed number of positions.