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.