2021 AMC 10A Spring 第 18 题

Question

设

f

是定义在正有理数集上的函数，并且

f(a\cdot b)=f(a)+f(b)

对所有正有理数

a

和

b

都成立。又假设

f

还满足：对每个质数都有

f(p)=p

，其中

p

为质数。下列哪个数

x

满足

f(x) < 0

？

Let

f

be a function defined on the set of positive rational numbers with the property that

f(a\cdot b)=f(a)+f(b)

for all positive rational numbers

a

and

b.

Suppose that

f

also has the property that

f(p)=p

for every prime number

p.

For which of the following numbers

x

is

f(x) < 0?

\dfrac{17}{32}

\dfrac{11}{16}

\dfrac{7}{9}

\dfrac{7}{6}

\dfrac{25}{11}

答案：E

知识点：函数方程质因数分解

难度评级：1280

小提示：

先确定

f(p^e)

和

f(\frac{a}{b})

的值。

First determine

f(p^e)

and

f(\frac{a}{b})

大提示：

用质因数分解逐个计算选项。

Evaluate the choices by prime factorization

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文字解答：

反复应用该函数方程可得：对每个质数

p

和正整数

e

，都有

f(p^e)=ef(p)=ep

。另外

f(a)=f\left(\frac ab\right)+f(b)\text{，}

所以

f(\frac{a}{b})=f(a)-f(b)

。

用质因数分解逐项计算各个选项，得

\begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1 \end{aligned}\text{。}

只有最后一个值是负数。

所以正确答案是 E。

Repeated use of the functional equation gives

f(p^e)=ef(p)=ep

for every prime

p

and positive integer

e.

Also,

f(a)=f\left(\frac ab\right)+f(b),

so

f(\frac{a}{b})=f(a)-f(b).

Evaluating the choices by prime factorization,

\begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1. \end{aligned}

Only the final value is negative.

Thus, E is the correct answer.