First, the number of trailing zero equals the number of divisor $5$ that $90!$ has: $$\left\lfloor{\frac{90}{5}}\right\rfloor+\left\lfloor{\frac{90}{5^2}}\right\rfloor=21$$

This means that $90!$ has $21$ trailing zeros. Let $N=\frac{90!}{10^{21}}$. Then the desired answer $n$ equals $N\pmod{100}$.

Clearly, $N$ still has more than two divisors of $2$. Hence, $N\equiv 0\pmod{4}$. In order to calculate $N\pmod{100}$, we just need to compute $N\pmod{25}$.

By the conclusion of # 4160, we know $(5k+1)(5k+2)(5k+3)(5k+4)\equiv -1\pmod{25}$. In order to use this conclusion, let $$M=1\times 2\times 3\times 4\times \underline{1} \times 6\times \cdots \times 86\times 87\times 88\times 89\times \underline{18}$$

That is, to eliminate all $5$ from $90!$, i.e. $5\rightarrow 1$, $10\rightarrow 2$, $\cdots$, $25\rightarrow 1$, $\cdots$, $90\rightarrow 18$. Then, we have $$\begin{array}{rl} M = &1\times 2 \times 3\times 4\times 1 \times 6\times \cdots \times 86\times 87\times 88\times 89\times 18 \\ =& (1\times 2\times 3\times 4)(6\times 7\times 8)\cdots(86 \times 87\times 88\times 89) \\ & (1\times 2\times 3\times 4)(6\times 7\times 8\times 9)\cdots(16\times 17\times 18) \\ & (1\times 2\times 3) \end{array} $$

The $2^{nd}$ last line above is corresponding to all the numbers which divides $5$, but not $25$. The last line is corresponding to those numbers which are multiple of $25$. Therefore $$M\equiv (-1)^{10}\times (-1)^3\times (16\times 17\times 19)(1\times 2\times 3)\equiv 24\pmod{25}$$

Meanwhile, we have $2^{21}\equiv (2^{10})^2\times 2\equiv (-1)^2\times  2\equiv 2 \pmod{25}$. It follows that $$N=\frac{M}{2^{21}}\equiv \frac{24}{2}\equiv 12\pmod{25}$$

Together with the fact of $N\equiv 0\pmod{4}$, we found $N\equiv \boxed{12}\pmod{100}$