How many consecutive zeros are there at the end of the integer obtained in the product 1² × 2⁴ × 3⁶ × 4⁸ × … × 25⁵⁰?
- (a) 50
- (b) 55
- (c) 100
- (d) 200 ✓ UPSC's answer
Why the answer is (d)
• The product is 1² × 2⁴ × 3⁶ × … × 25⁵⁰, i.e. each number k is raised to the power 2k.
• Trailing zeros come from pairs of 2 and 5; 2s are far more plentiful, so count the 5s.
• Only 5, 10, 15, 20 and 25 contain 5. Their powers are 10, 20, 30, 40 and 50.
• 5, 10, 15, 20 each contain one 5: 10 + 20 + 30 + 40 = 100 fives. 25 = 5² contains two, so 25⁵⁰ gives 2 × 50 = 100 fives.
• Total fives = 200, so the product ends in 200 zeros — option (d).
Why the other options are wrong
- (a) 50
- 50 is far too low: the powers of 5, 10, 15 and 20 alone give 10 + 20 + 30 + 40 = 100 fives, and 25⁵⁰ adds another 100.
- (b) 55
- 55 is too low: counting every factor of 5 gives 100 from 5, 10, 15, 20 and another 100 from 25⁵⁰, i.e. 200.
- (c) 100
- 100 counts only the fives from 5¹⁰, 10²⁰, 15³⁰ and 20⁴⁰ (or only one 5 per power of 25); 25⁵⁰ = 5¹⁰⁰ adds 100 more, giving 200.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2024, held on 16 June 2024. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.