# 1960 AHSME Problems/Problem 17

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## Problem 17

The formula $N=8 \times 10^{8} \times x^{-3/2}$ gives, for a certain group, the number of individuals whose income exceeds $x$ dollars. The lowest income, in dollars, of the wealthiest $800$ individuals is at least: $\textbf{(A)}\ 10^4\qquad \textbf{(B)}\ 10^6\qquad \textbf{(C)}\ 10^8\qquad \textbf{(D)}\ 10^{12} \qquad \textbf{(E)}\ 10^{16}$

## Solution

Plug $800$ for $N$ because $800$ is the number of people who has at least $x$ dollars. $$800 = 8 \times 10^{8} \times x^{-3/2}$$ $$10^{-6} = x^{-3/2}$$ In order to undo raising to the $-\frac{3}{2}$ power, raise both sides to the $-\frac{2}{3}$ power. $$(10^{-6})^{-2/3} = (x^{-3/2})^{-2/3}$$ $$x = 10^4 \text{ dollars}$$

The answer is $\boxed{\textbf{(A)}}$.

## See Also

 1960 AHSME (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 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
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