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

Alternate finish: multiply the denominator by 9 and notice that it is 1 less than $10^{10}$ . So the answer is very very close to $\boxed{\mathrm{(B) } 9}$ .

~JH. L

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2011 AMC 10B Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See Also