1981 AMC 12 第 13 题

Question

假设在每年年末，一个货币单位相较该年年初会损失其价值的

10\%

。求最小的整数

n

，使得经过

n

年后，这个货币单位至少损失其价值的

90\%

。（四舍五入到小数点后三位，

\log_{10}3

为

0.477

。）

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 unit of money will have lost at least

90\%

of its value. (To the nearest thousandth

\log_{10}3

is

0.477.

)

14

16

18

20

22

经过

n

年后，剩余价值所占的比例为

0.9^n

After

n

years, the fraction of value remaining is

0.9^n

利用

\log_{10}0.9=2\log_{10}3-1

Use

\log_{10}0.9=2\log_{10}3-1

至少损失

90\%

意味着

0.9^n\le0.1

。现在

\begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046 \end{aligned}\text{。}

因此

n(-0.046)\le-1

，所以

n\ge21.739\ldots

。最小的整数是

22

。

所以正确答案是 E。

At least

90\%

lost means

0.9^n\le0.1.

Now

\begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046. \end{aligned}

Thus

n(-0.046)\le-1,

so

n\ge21.739\ldots.

The least integer is

22.

Therefore, the correct answer is E.

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