Task B-1.1. How many zeros does the number $N=x^{3}+3 x^{2}-9 x-27$ end with, if $x=999997$?
## First solution.

We can write the expression in the form

$$
N=x^{3}+3 x^{2}-9 x-27=x^{2}(x+3)-9(x+3)=(x+3)\left(x^{2}-9\right)=(x+3)^{2}(x-3) \text { (3 points) }
$$

Then for $x=999997$ we get

$$
N=(999997+3)^{2}(999997-3)=10^{12} \cdot 999994
$$

which means that the number $\mathrm{N}$ ends with $\mathrm{12}$ zeros.