What's more likely: $7$-digit number with no $1$'s or at least one $1$ among its digits? 
A $7$-digit number is chosen at random. Which is more likely: the number has no $1$'s among its digits or the number has at least one $1$ among its digits?

Here's how I did it: The question is asking whether $8(9)^6$ (the number of those with no $1$'s among its digits) or $9(10)^6 - 8(9)^6$ (the number of those with at least one $1$ among its digits). Some tedious multiplying shows that $8(9)^6 = 4241528 < 4500000$, which demonstrates that $9(10^6) - 8(9)^6$ i.e. the number having at least one $1$'s among its digits is more likely.
However, I am wondering if there is a slicker way to get the answer without having to do any tedious multipication.
 A: There is a more tedious way to do this. We have that
$$
9\cdot (10)^6- 8\cdot(9)^6\geq 8\cdot (9)^6
$$
if and only if
$$
16\cdot 9^6\leq 9\cdot 10^6.
$$
Now take the square root on both sides to get that the first inequality holds if and only if
$$
4\cdot 9^3\leq 3\cdot 10^3 \Leftrightarrow \\ 
\frac43\leq \left( \frac{10}{9}\right)^3\Leftrightarrow\\
0.75= \frac34\geq \left( \frac{9}{10}\right)^3= \frac{729}{1000}= 0.729.
$$
A: Note that $8\cdot 9^6 < 9\cdot 10^6-8\cdot 9^6$ is equivalent to
$$2\cdot 8\cdot 9^6 < 9\cdot 10^6.$$
Getting rid of common factors on both sides by considering prime factorizations, we can reduce this as far as
$$3^{10} < 2^2 \cdot 5^6.$$
Now both sides are perfect squares, so by taking square roots this is equivalent to
$$3^5 < 2 \cdot 5^3.$$
This we might actually be able to calculate:
\begin{align}
3^5 &= 3\cdot 9^2 = 3\cdot 81 = 243, \\
2\cdot 5^3 &= 2\cdot 5\cdot 25 = 10\cdot 25 = 250.
\end{align}
This confirms the original inequality.
A: We want to show that $$ \frac{8·9^6}{9·10^6} \ = \ \frac{8·9^6}{10 \ · \ 9·10^5} \ = \ 0.8 · (0.9)^5 \ < \ \frac12 \ \ . $$
Since $ \ 0.9^5 \ = \ (1 - 0.1)^5 \ = \ 1 - 5·(0.1) + 10·(0.1)^2 - \ldots \ < \ 0.61 \ \ ,  $  we have  $ \  0.8 · (0.9)^5 \ < \ 0.8 · 0.61 \ < \ 0.5 \ \ . $  (The exact ratio you sought is $ \ \frac{4251528}{9000000} \ = \ 0.472392 \ \ . ) $
Thus, the set of seven-digit numbers with no 1's is the smaller of the two sets.
A: If you remember a couple of common $\log$ values it can be done easily. $\log 2\approx 0.301<0.302$ and $\log 3\approx 0.4771<0.4772$. So, $$4\log 2+10\log 3<1.208+4.772=5.98<6$$
This means that, taking antilog, $16\cdot (9)^5<10^6$ so that $8\cdot (9)^6<9\cdot (10)^6-8\cdot (9)^6$.
