Binary of 1729 decimal – UGC NET Question

Converting 1729 to binary by repeated division by 2 gives the remainder sequence that reconstructs as 11011000001₂. Expanding 11011000001₂ as 1×1024 + 1×512 + 0×256 + 1×128 + 1×64 + 0×32 + 0×16 + 0×8 + 0×4 + 0×2 + 1×1 yields 1729. This confirms that the binary string exactly matches the decimal number. Thus option A is the correct representation.

Option A:
11011000001₂ uses bits set in positions corresponding to 1024, 512, 128, 64 and 1. Adding these values gives 1729, which matches the decimal number. The presence of zero bits in all other positions preserves this sum exactly, making this option correct.

Option B:
11011100001₂ includes an additional 256 contribution compared to the correct pattern because of the extra 1 in a higher position. Its expansion becomes 1024 + 512 + 256 + 64 + 1 = 1857, which is larger than 1729. Hence this pattern is not correct.

Option C:
11011000010₂ shifts one of the 1 bits to the 2¹ position instead of the 2⁰ position. That makes the total 1024 + 512 + 128 + 64 + 2 = 1730. Because 1730 is not equal to 1729, this option is numerically incorrect.

Option D:
10111000001₂ uses 1024 + 256 + 128 + 64 + 1 = 1473 as its decimal value. This is significantly smaller than 1729, showing that its pattern of ones in high-order bits differs from the correct one. Therefore it is not the correct representation for 1729.