UPSC Prelims 2026 · CSAT
Question 66: Basic Numeracy
A is a 2-digit number with different digits. B is also a 2-digit number and is obtained by reversing the digits of A. If A − B is a multiple of 27, where A > B, how many such different A’s are possible?
AnswerUPSC’s answer is (b).
Let A = 10a + b and B = 10b + a, so A − B = 9(a − b). For a multiple of 27, a − b must be a multiple of 3, with a > b. B is also two-digit, so b cannot be 0. a − b = 3 gives (a, b) = (4,1), (5,2), (6,3), (7,4), (8,5), (9,6): 6 numbers. a − b = 6 gives (7,1), (8,2), (9,3): 3 numbers. a − b = 9 would need b = 0, which is not allowed. Total 9.
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: 6 counts only the digit difference of 3.
- (b) Right: 6 + 3 = 9.
- (c) Wrong: 12 wrongly allows b = 0, where B would not be two-digit.
- (d) Wrong: 18 overcounts.
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.