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Difference between revisions of "2011 AMC 10B Problems/Problem 10"

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<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math>
 
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101 </math>
  
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== Solution ==
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== Solution 1 ==
  
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The sum of the other ten elements is the same as ten <math>1</math>s. <math>10^{10}</math> is the same as <math>1</math> followed by ten <math>0</math>s. If you subtract one, it is equal to ten <math>9</math>s. Therefore if you divide the sum of the other ten elements by the largest element, it is closest to <math>\boxed{\mathrm{(B) \ } 9}</math>
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The requested ratio is <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Using the formula for a geometric series, we have <cmath>10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},</cmath> which is very close to <math>\dfrac{10^{10}}{9},</math> so the ratio is very close to <math>\boxed{\mathrm{(B) \ } 9}.</math>
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== Solution 2 ==
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The problem asks for the value of <cmath>\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.</cmath> Written in base 10, we can find the value of <math>10^9 + 10^8 + \ldots + 10 + 1</math> to be <math>1111111111.</math> Long division gives us the answer to be <math>\boxed{\mathrm{(B) \ } 9}.</math>
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Alternate finish: multiply the denominator by 9 and notice that it is 1 less than <math>10^{10}</math>. So the answer is very very close to <math>\boxed{\mathrm{(B) } 9}</math>.
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~JH. L
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== Solution 3 ==
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Let <math>f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}</math>. We are approximating <math>f(10)</math>. Trying several small values of <math>n</math> gives answers very close to <math>9</math>, so our answer is <math>\boxed{\textbf{(B)}~9}</math>. Note that <math>f(1)=10</math>, but <math>f(2)=\dfrac{100}{11}\approx9.09.</math>  
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~Technodoggo
  
 
== See Also==
 
== See Also==
  
 
{{AMC10 box|year=2011|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2011|ab=B|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 21:43, 31 August 2023

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

Solution 3

Let $f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}$. We are approximating $f(10)$. Trying several small values of $n$ gives answers very close to $9$, so our answer is $\boxed{\textbf{(B)}~9}$. Note that $f(1)=10$, but $f(2)=\dfrac{100}{11}\approx9.09.$ ~Technodoggo

See Also

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