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NDA 2 2025 Maths, question 4

Topic: Algebra

Q4

Let nn be a natural number. The number of consecutive zeros at the end of the expansion of n!n! is exactly 22. How many values of nn are possible?

  1. (a) 33
  2. (b) 44
  3. (c) 55Correct answer
  4. (d) More than 5

Solution.

Answer: (c) 55

  1. Each trailing zero needs a factor 10, and factors 5 are limiting.
  2. Exactly two factors 5 occur for n=10,11,12,13,14.
  3. At n=15 there are already three.
  4. Thus five values are possible.