How many zeroes are there at the end of the following product?
1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60
Answer & explanation
Answer: (a) 10
Each trailing zero needs one pair of 2 and 5. Writing the product as 5¹² × 12! shows 14 fives but only 10 twos, so the twos run out first and there are 10 zeroes.
- The terms after 1 are 5 × 1, 5 × 2, …, 5 × 12, so the product = 5¹² × (1 × 2 × … × 12) = 5¹² × 12!.
- Fives: 12 from 5¹², plus 2 in 12! (from 5 and 10) = 14.
- Twos: only from 12! — 6 + 3 + 1 = 10 (multiples of 2, 4 and 8 up to 12).
- Zeroes = min(10, 14) = 10.
- Check: 12! = 479001600 has two 5s and ten 2s, as used.
Remember · Trailing zeroes = the smaller of the counts of 2s and 5s. Do not assume 5s are always scarcer.
Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·