1981 AHSME Problems/Problem 13

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

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1981 AHSME Problems/Problem 13

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