How many consecutive zeros are there at the end of the integer obtained in the product 1² × 2⁴ × 3⁶ × 4⁸ × ⋯ × 25⁵⁰?
Answer & explanation
Answer: (d) 200
The product is n raised to the power 2n for n = 1 to 25. Twos are plentiful, so count the fives: only 5, 10, 15, 20 and 25 contribute, giving 10 + 20 + 30 + 40 + 100 = 200.
- Each zero at the end needs one 2 and one 5; there are far more 2s than 5s here, so count the 5s.
- Only multiples of 5 carry a 5: 5¹⁰ gives 10 fives, 10²⁰ gives 20, 15³⁰ gives 30, 20⁴⁰ gives 40.
- 25 = 5², so 25⁵⁰ = 5¹⁰⁰ gives 100 fives.
- Total fives = 10 + 20 + 30 + 40 + 100 = 200, so there are 200 zeros at the end.
- Check: forgetting that 25 holds two 5s gives 150, which is not even an option.
Remember · Trailing zeros = number of 5s (2s are always more). Remember 25 = 5², so its power counts double.
Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·