2026 CSAT, all questions
UPSC Prelims 2026 · CSAT
Question 42: Basic Numeracy
The digit in the unit place of the number 6^129 × 7^307 is
a2
b4
c8
d6
AnswerUPSC’s answer is (c).
Any power of 6 ends in 6. Powers of 7 end in 7, 9, 3, 1 in a repeating cycle of four. Since 307 leaves remainder 3 when divided by 4, 7^307 ends in 3. The product therefore ends in the units digit of 6 × 3 = 18, which is 8.
Why each option is right or wrong
- (a) Wrong: 6 × 3 = 18 ends in 8, not 2.
- (b) Wrong: the units digit is 8, not 4.
- (c) Right: 6 × 3 = 18, units digit 8.
- (d) Wrong: 6 is the units digit of 6^129 alone; 7^307 changes it.
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.