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.
 A: Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] <\frac{2}{9}\left(m!\right)^3.$$
So you have to go at least far enough so that
$$
\frac{2}{9}\left(m!\right)^3>10^{6},
$$
or $m! > \sqrt[3]{4500000} > 150$.  So $m=5$ (corresponding to $n=15$) isn't far enough; the smallest $n$ satisfying your inequality will be at least $16$.
Similarly, for $n=3m+1$,
$$
3^{-3m-1}(3m+1)!=\left[\left(m+\frac{1}{3}\right)m\left(m-\frac{1}{3}\right)\right]\cdots \left[\frac{4}{3}\cdot1\cdot\frac{2}{3}\right]\cdot\frac{1}{3} < \frac{1}{3}(m!)^3,$$
so you need $m!>\sqrt[3]{3000000}> 140$, and $m=5$ (that is, $n=16$) is still too small.
Finally, for $n=3m+2$,
$$
3^{-3m-2}(3m+2)!=\left[\left(m+\frac{2}{3}\right)\left(m+\frac{1}{3}\right)m\right]\cdots \left[\frac{5}{3}\cdot\frac{4}{3}\cdot1\right]\cdot\frac{2}{3}\cdot\frac{1}{3} > \frac{560}{729}(m!)^3,
$$
where the coefficient comes from the last eight terms, so it is sufficient that $m! > 100\cdot\sqrt[3]{729/560}.$  To show that $m=5$ is large enough, we need to verify that $(12/10)^3=216/125 > 729/560$.  Carrying out the cross-multiplication, you can check without a calculator that $216\cdot 560 =120960$ is larger than $729\cdot 125=91125$, and conclude that $m=5$ (that is, $n=17$) is large enough.  The inequality therefore holds for exactly all $n\ge 17$.
A: I would use Stirling's approximation $n!\approx \frac {n^n}{e^n}\sqrt{2 \pi n}$ to get $\left( \frac {3e}n\right)^n \sqrt{2 \pi n} \lt 10^{-6}$.  Then for a first cut, ignore the square root part an set $3e \approx 8$ so we have $\left( \frac 8n \right)^n \lt 10^{-6}$.  Now take the base $10$ log asnd get   $n(\log 8 - \log n) \lt -6$  Knowing that $\log 2 \approx 0.3$, it looks like $16$ will not quite work, as this will become $16(-0.3)\lt 6$.  Each increment of $n$ lowers it by a factor $5$ around here, or a log of $0.7$.  We need a couple of those, so I would look for $18$.
Added:  the square root I ignored is worth a factor of $10$, which is what makes $17$ good enough.a
Alpha shows that $17$ is good enough.
A: 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".
A: how about make a function?$f(n)=\frac{3^n}{n!}-10^{-6}$ or maybe $f(n)=\frac{3^n}{n!10^{-6}}$
