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UPSC CSE CSAT 2020 · Paper II (CSAT) · Question 7 · Number system

गुणनफल 1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60 के अंत में कितने शून्य होंगे?

UPSC Civil Services (Preliminary) Examination 2020, General Studies Paper II (CSAT), Question 7 — Number system.

CSAT 2020 · Q7

Number system Medium

गुणनफल

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

के अंत में कितने शून्य होंगे?

Read this question in English

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

(a) 10 · (b) 12 · (c) 14 · (d) 15

उत्तर और व्याख्या · explanation in English

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.

  1. The terms after 1 are 5 × 1, 5 × 2, …, 5 × 12, so the product = 5¹² × (1 × 2 × … × 12) = 5¹² × 12!.
  2. Fives: 12 from 5¹², plus 2 in 12! (from 5 and 10) = 14.
  3. Twos: only from 12! — 6 + 3 + 1 = 10 (multiples of 2, 4 and 8 up to 12).
  4. Zeroes = min(10, 14) = 10.
  5. 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). ·

Cite this question: UPSC CSE CSAT 2020, Q7 — minimalistias.com/pyqs/csat/2020/q7

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