Let n be a natural number. The number of consecutive zeros at the...

The number of consecutive zeros at the end of $n!$ is determined by the exponent of $5$ in the prime factorization of $n!$.

Using Legendre's formula, the exponent of $5 in n!$ is given by:

$E_5(n!) = \left\lfloor \dfrac{n}{5} \right\rfloor + \left\lfloor \dfrac{n}{5^2} \right\rfloor + \left\lfloor \dfrac{n}{5^3} \right\rfloor + \dots$

Given that $E_5(n!) = 2$, $n$ must be less than $25$, which means \left\lfloor \dfrac{n}{25} \right\rfloor = 0$and all higher terms are also$0$.<br><br>Thus, the equation simplifies to:<br><br>\left\lfloor \dfrac{n}{5} \right\rfloor = 2$

This inequality holds for:

$2 \le \dfrac{n}{5} \lt 3$

$10 \le n \lt 15$

Since $n$ is a natural number, the possible values of $n$ are $10, 11, 12, 13, 14$.

There are exactly $5$ possible values for $n$.

Answer: $5$