# 2011 AMC 10B Problems/Problem 10

## Problem

Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer? $\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

## Solution 1

The requested ratio is $$\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.$$ Using the formula for a geometric series, we have $$10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},$$ which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{\mathrm{(B) \ } 9}.$

## Solution 2

The problem asks for the value of $$\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.$$ Written in base 10, we can find the value of $10^9 + 10^8 + \ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\boxed{\mathrm{(B) \ } 9}.$

## See Also

 2011 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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