# For what values of $n$ is $10^8 < n! < 10^{12}$?

This question is from the introduction of a handbook on combinatorics. So far, the material covered contains:

• counting problems solved by using trees, Pascal's triangle,...
• definition of $$n$$ factorial
• definitions of combinations, variations (based on allowing repitition and if order matters).

No formulas, except for the formula for $$n$$ factorial have been seen.

The question then asks to find all values of $$n$$ such that $$10^8 < n! < 10^{12}$$ and the given answers states that this holds for $$n = 12,13,14$$ (only a numerical answer is given, no reasoning).

I tried solving this by taking the logarithm with base 10 from all sides, but this did not really help (seemed to brute force to me).

question: is there some 'nice' way to prove this answer?

For the lower bound, observe that $$10^7=\underbrace{10\cdot 10\cdot ....\cdot 10}_{7\text{ times}}>10\cdot9\cdot8\cdot7\cdot6\cdot4\cdot3=9!\implies 10^8>10\cdot9!=10!$$

Thus $$n>10$$. Once noticed that $$12!$$ works (which isn’t very hard) observe the following for the upper bound

if $$a\in(10^8, 10^{12})$$, then $$10^4\cdot a\not\in (10^8, 10^{12})$$

In particular $$a\cdot 13\cdot 14\cdot 15\cdot 16\not\in (10^8, 10^{12})$$ since $$a\cdot 13\cdot 14\cdot 15\cdot 16>a\cdot 10^4$$.

Hence, since $$12!$$ satisfies the inequality, $$12!\cdot 13\cdot 14\cdot 15\cdot 16=16!$$ doesn’t.

• Nice answer for the lower bound. I'll try to figure out if a similar reasoning gives an upper bound! – Student Apr 7 at 13:54
• I’ve edited the answer for the upper bound ;) – Dr. Mathva Apr 7 at 14:15
• Thansk for the edit, but 11! Does not satisfy this inequality. – Student Apr 7 at 14:20
• I know, but the upper and lower bound aren’t necessarily the lowest and the uppest hounds, but the best ones I managed to achieve – Dr. Mathva Apr 7 at 14:29
• Thank you. Not sure if this is the intended answer, but I like this one very much :) – Student Apr 7 at 14:33

There's a nice way (called "Stirling's approximation") to get a narrower estimate, but I think that the goal here was to have you think about things. For instance, you know that $$n > 8$$, because at $$n = 8$$, the left hand side is $$10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10 \cdot 10$$ while the middle is $$1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8$$ Since each factor of the second is smaller than the corresponding factor of the first, you clearly need a larger $$n$$.

On the other hand, $$21!$$ has $$12$$ factors that are at least $$10$$, namely $$10, 11, 12, \ldots, 21$$ (as well as a bunch of smaller factors). So $$21!$$ is definitely bigger than $$10^{12}$$. At this point, you know that $$n$$ is between $$8$$ and $$21$$, and trying a few numbers in between narrows things down pretty fast.

But along the way, I expect the author hopes that you've learned something about estimating factorials by not-very-subtle methods. (The proof I know of Stirling's formula, for instance, involves integrals.)

• This seems like what was intented. This answer only uses what is assumed to be known, so I don't think that things can become much better. Thank you! – Student Apr 7 at 13:53

By taking logs of both sides, $$8\ln{(10)}\lt\ln{(n!)}\lt12\ln{(10)}$$ Then applying Stirling's approximation we have $$8\ln{(10)}\lt n\ln{(n)}-n+\frac12\ln{(2\pi n)}\lt12\ln{(10)}$$ $$18.4...\lt n\ln{(n)}-n+\frac12\ln{(2\pi n)}\lt27.6...$$ Letting $$f(x)=x\ln{(x)}-x+\frac12\ln{(2\pi x)}$$ we have that $$f(11)=17.4...$$ $$f(12)=19.9...$$ $$f(13)=22.5...$$ $$f(14)=25.1...$$ $$f(15)=27.8...$$ So the values of $$n$$ for which this is valid are $$12,13$$ and $$14$$.

• Thanks for your answer. I suspect that the authors of this handbook did not intented to use this approximation (book has as audience 17/18 y/o, and gives an introduction to combinatorics). I never heard of this approximation, so I still find this answer very interesting, so thanks again! – Student Apr 7 at 13:52