How many three-digit numbers can be expressed as an integral power of 2?
- (a) 1
- (b) 2
- (c) 3 ✓ UPSC's answer
- (d) 4
Why the answer is (c)
• A three-digit number must be greater than or equal to 100 and less than or equal to 999.
• We need to find integers $n$ such that $100 \le 2^n \le 999$.
• Calculate powers of 2: $2^6 = 64$ (too small), $2^7 = 128$ (valid), $2^8 = 256$ (valid), $2^9 = 512$ (valid), $2^{10} = 1024$ (too large).
• The valid powers are $2^7$, $2^8$, and $2^9$.
• There are exactly 3 such numbers, so the correct option is (c).
Why the other options are wrong
- (a) 1
- Option (a) is incorrect because there are three valid numbers (128, 256, 512), not just one.
- (b) 2
- Option (b) is incorrect because it misses one of the three valid powers of 2 within the three-digit range.
- (d) 4
- Option (d) is incorrect because $2^{10} = 1024$ is a four-digit number, so there are only three valid three-digit powers.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.