UPSC Prelims 2024 · CSAT
Question 9: Basic Numeracy
How many consecutive zeros are there at the end of the integer obtained in the product 1^2 × 2^4 × 3^6 × 4^8 × ··· × 25^50?
AnswerUPSC’s answer is (d).
The product is the product of n^(2n) for n = 1 to 25. Trailing zeros equal the smaller of the powers of 2 and of 5. Count the 5s: n = 5, 10, 15, 20 give 2n fives each, i.e. 10, 20, 30, 40, and n = 25 gives 2 × 25 × 2 = 100 because 25 = 5². Total 200. Twos are far more numerous (16^32 alone gives 128), so the 5s limit the zeros: 200.
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.
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. 50 ignores that each n is raised to 2n and that 25 holds two fives.
- (b) Wrong. 55 is not what the count of fives gives; the total is 200.
- (c) Wrong. 100 counts only the fives from 25^50 and misses those from 5, 10, 15 and 20.
- (d) Right. Fives total 10 + 20 + 30 + 40 + 100 = 200, fewer than the twos.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2024, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.