2011 AMC 10B Problem 10

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 $\boxed{\textbf{(B)}\ 9}$.

A geometric series with ratio $10$ is dominated by its last term, and the sum of all earlier terms is roughly the last term divided by $r - 1 = 9$. The exact repunit formula $\frac{10^n - 1}{9}$ makes the estimate rigorous.

Compare $10^{10} = 10{,}000{,}000{,}000$ with $1{,}111{,}111{,}111$: nine copies of the sum give $9{,}999{,}999{,}999$, which is $1$ short of $10^{10}$, so the ratio is just above $9$.

• Comparing $10^{10}$ only with the next largest term $10^9$ and answering $10$.
• Counting the largest element in the sum as well, which pushes the estimate toward $1$ or $10$.