Question

Find the number of trailing zeroes in (2⁵ × 5²)!

1. 2
2. 192
3. 199
4. 7

Answer 1

Correct Answer - Option 3 : 199

Given:

(2⁵ × 5²)!

Concept Used:

n! = n(n – 1)(n – 2)…1

Number of zeros = Power of 5

In case of factorial the power of 5 is the limiting factor because 5 is less likely to occur than 2.

Maximum power of 5 in n! = n/5 + n/5² + n/5³ +... (Consider integer values only)
Calculation:

(2⁵ × 5²)! = 800! = 800 × 799 × … × 1

Maximum power of 5 in 800! = 800/5 + 800/5² + 800/5³ + 800/5⁴ = 160 + 32 + 6 + 1 = 199

Here we consider integer values only and ignore the remainder.

∴ The number of zeros is 199.