UPSC Prelims 2026 · CSAT
Question 45: Basic Numeracy
How many three-digit numbers can be expressed as an integral power of 2?
AnswerUPSC’s answer is (c).
2^6 = 64 has two digits, 2^7 = 128 is the first three-digit power, and 2^10 = 1024 already has four digits. So the three-digit powers are 2^7 = 128, 2^8 = 256 and 2^9 = 512: exactly three. Doubling 512 gives 1024, which passes 999, so no further power qualifies.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
This follows UPSC’s provisional key; the final key could change it.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) Wrong: 128, 256 and 512 all qualify.
- (b) Wrong: three powers of 2 lie between 100 and 999.
- (c) Right: 128, 256 and 512.
- (d) Wrong: 1024 has four digits, so there is no fourth.
Only the provisional key (27 May 2026) was found; the final key replaces it when found.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2026, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.