2011 AMC 10B Problem 10

1. The largest number is $10^{10} = 10{,}000{,}000{,}000$.
2. The other ten numbers are $1, 10, 100, \ldots, 10^9$. Adding them, each power of $10$ puts a $1$ in a different place:
$$1 + 10 + 100 + \cdots + 1{,}000{,}000{,}000 = 1{,}111{,}111{,}111.$$
3. We want $10{,}000{,}000{,}000 \div 1{,}111{,}111{,}111$. Try $9$:
$$9 \times 1{,}111{,}111{,}111 = 9{,}999{,}999{,}999.$$
4. That is only $1$ less than $10{,}000{,}000{,}000$. So the ratio is a tiny bit more than $9$, about $9.000000009$.
5. The closest whole number is $9$.

The answer is $\boxed{\textbf{(B)}\ 9}$.

When each number is $10$ times the one before, the last number is bigger than all the others put together, but not by a factor of $10$: the earlier numbers add up to about one ninth of the last one. The quick check with $1{,}111{,}111{,}111 \times 9$ shows it without any formulas.

Try it on a smaller set first: for $\{1, 10, 100\}$ the ratio is $100 \div 11 \approx 9.1$; for $\{1, 10, 100, 1000\}$ it is $1000 \div 111 \approx 9.01$. The ratio gets closer and closer to $9$ as the set grows, so for eleven numbers it is essentially $9$.

The ten smaller elements sum to
$$1 + 10 + 10^2 + \cdots + 10^9 = 1{,}111{,}111{,}111 = \frac{10^{10} - 1}{9}.$$

The requested ratio is
$$\frac{10^{10}}{(10^{10} - 1)/9} = \frac{9 \cdot 10^{10}}{10^{10} - 1},$$
which is slightly more than $9$ (the denominator is a hair less than $10^{10}$) and far closer to $9$ than to $10$. Numerically, $10^{10}/1{,}111{,}111{,}111 \approx 9.000000009$.

The answer is $\boxed{\textbf{(B)}\ 9}$.