1960 AMC 12 Problem 17

Attempt Problem 17 of the 1960 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1960 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

The formula

N=8\cdot10^8\cdot x^{-\frac{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:

10^4

10^6

10^8

10^{12}

10^{16}

Set

N=800

at the income cutoff

After simplifying powers of

10,

solve

x^{\frac{3}{2}}=10^6

At the cutoff,

800=8\cdot10^8x^{-\frac{3}{2}}.

Hence

x^{\frac{3}{2}}=10^6,

and raising both sides to the

\frac{2}{3}

power gives

x=10^4.

Thus, the correct answer is A.

Problem 17 in other years