The expression $$1\times 2\times 3\times \cdots \times n$$ has exactly $$128$$ consecutive zeros at the end of its result, then the maximum value of $$n$$ is~\uline{~~~~~~~~~~}~． 
Overseas Competition->Knowledge Point->Number Theory Modules
There are $$128$$ zeros at the end of the result, and the estimated answer is close to $$500$$.  $$\left[ \frac{500}{5} \right]+\left[ \frac{500}{25} \right]+\left[ \frac{500 }{125} \right]=100+20+4=124$$,  Just add $$4$$ numbers including $$5$$: $$505$$, $$510$$, $$515$$, $$520$$, so the maximum value of $$n$$ that satisfies the condition is $$524$$ . 