<s>[INST] 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.
 [/INST] 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.
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