(Q). If p = x¹⁰⁰⁰(1000!) & p is divisible by 10¹⁰⁰⁰. then what would be the minimum number of zeros at the end of p.
Solution:-
first find number of trailing zeros in 1000!
Hence 1000! has 249 trailing zeros.
∴ 1000! = 2⁹⁹⁴ × 5²⁴⁹ × (other factors)
now 10¹⁰⁰⁰ has 1000 zeros after 1 (i.e. 1000 trailing zeros) & 1000! has 249 trailing zeros.
So for p to be completely divisible by 10¹⁰⁰⁰ , remaining zeros should come from “x¹⁰⁰⁰” term & x¹⁰⁰⁰ will have trailing zeros only when x is a multiple of 2 & 5.
So to find minimum zeros at the end of p take the least value of x as 2×5
x = 2×5 = 10
∴ p should have 1249 zeros at the end Answer