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Difference between revisions of "1981 AHSME Problems/Problem 13"

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==Problem==
 
==Problem==
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Suppose that at the end of any year, a unit of money has lost 10% of the value it had at the beginning of that year. Find the smallest integer <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>\log_{10}{3}=0.477</math>).
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Suppose that at the end of any year, a unit of money has lost <math>10\%</math> of the value it had at the beginning of that year. Find the smallest integer <math>n</math> such that after <math>n</math> years, the money will have lost at least <math>90\%</math> of its value (To the nearest thousandth <math>\log_{10}{3}=0.477</math>).
  
 
<math>\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22</math>
 
<math>\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22</math>
  
 
==Solution==
 
==Solution==
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What we are trying to solve is <math>\log_{0.9}{0.1}=n</math>. This turns into <math>\frac{\log{0.1}}{\log{0.9}}=\frac{-1}{\log{9}-1}=n</math> We know that <math>\log_{10}{3}=0.477</math>, thus by log rules we have <math>2\log_{10}{3}=\log_{10}{9}=2*0.477=0.954</math>, thus <math>n=\frac{1}{.046} \approx 21.7</math>, and our answer is <math>\boxed{(\text{E}) 22}</math>.
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After <math>n</math> years, the money will be worth <math>(0.9)^n</math> times what it is currently worth. Therefore, the goal is to find the smallest integer such that <math>(0.9)^n < 0.1</math>.  Taking the base-<math>10</math> logarithm of both sides yields <math>n \log_{10}{0.9} < \log_{10}{0.1}</math>, which is equivalent to <math>n > \frac{\log_{10}{0.1}}{\log_{10}{0.9}} = \frac{-1}{-1+2\log_{10}{3}}</math>.  
  
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-edited by Maxxie and maxamc
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Substituting <math>\log_{10}{3} \approx 0.477</math> yields <math>n > \frac{-1}{-1 + 2 \cdot 0.477} = \frac{-1}{-0.046} \approx 21.7</math>, so the minimum possible integer value of <math>n</math> is <math>\boxed{(\textbf{E})\ 22}</math>.
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-edited by Maxxie, maxamc, j314andrews
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==See also==
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{{AHSME box|year=1981|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 17:06, 18 August 2025

Problem

Suppose that at the end of any year, a unit of money has lost $10\%$ of the value it had at the beginning of that year. Find the smallest integer $n$ such that after $n$ years, the money will have lost at least $90\%$ of its value (To the nearest thousandth $\log_{10}{3}=0.477$).

$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 22$

Solution

After $n$ years, the money will be worth $(0.9)^n$ times what it is currently worth. Therefore, the goal is to find the smallest integer such that $(0.9)^n < 0.1$. Taking the base-$10$ logarithm of both sides yields $n \log_{10}{0.9} < \log_{10}{0.1}$, which is equivalent to $n > \frac{\log_{10}{0.1}}{\log_{10}{0.9}} = \frac{-1}{-1+2\log_{10}{3}}$.

Substituting $\log_{10}{3} \approx 0.477$ yields $n > \frac{-1}{-1 + 2 \cdot 0.477} = \frac{-1}{-0.046} \approx 21.7$, so the minimum possible integer value of $n$ is $\boxed{(\textbf{E})\ 22}$.

-edited by Maxxie, maxamc, j314andrews

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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