Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$.
Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$
And I appreciate hint rather than explicit solution.
Thank You.

How about we overestimate $3^n$ as $\sqrt{10}^n$ and underestimate every contribution to the factorial beyond $10$ as only $10$?  Then $$\frac{3^n}{n!} \le \frac{10^5 \cdot 10^{\frac{n-10}2}}{10! \cdot 10^{n-10}} $$ Since $10!$ is about $10^{6.5}$ we have $$ 10^{-1.5} \cdot 10^{-\frac{n-10}2} \le 10^{-6} $$ After taking the $\log_{10}$ we have $$ -1.5 -\frac{n-10}2 \le -6$$ 
$$ 1.5 + \frac{n-10}2 \ge 6$$ 
$$ \frac{n-10}2 \ge 4.5$$ 
$$ n-10 \ge 9 $$
$$ n \ge 19$$
Since you asked for a solution and not the minimal solution this should suffice, and is more plausibly done without a calculator.  The only real mental calculation was remembering the approximate value of $10!$ (well, and $3^2$, but let's be reasonable).
Note that if you know the factorials for higher numbers (or exact values for some powers of 3, such as $3^7 = 2187$) then you could refine this argument using higher $n$ for the "fixed part".
