2021 AMC 12A Spring 第 18 题

Question

设

f

是定义在正有理数集合上的函数，对所有正有理数

a

和

b

都满足

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

。又假设

f

对每个质数

p

都有

f(p) = p

。下列哪个数

x

满足

f(x) \lt 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) \lt 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：1950

小提示：

这个规则使

f

对质因数可加，且

f(\frac{1}{p}) = -f(p) = -p

The rule makes

f

additive over prime factors, and

f(\frac{1}{p}) = -f(p) = -p

大提示：

对

x = \prod p^{e_p}

，计算

f(x) = \sum e_p\, p

（分母中的质数对应负指数）

For

x = \prod p^{e_p},

compute

f(x) = \sum e_p\, p

(with negative exponents for denominators)

解答：

函数方程使

f

完全可加：若

x = \prod p^{e_p}

，则

f(x) = \sum e_p\, f(p) = \sum e_p\, p

，其中分母中的质数贡献负指数，因为

f(\frac{1}{p}) = -p

。

逐项计算：

f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7

，

f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3

，

f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1

，

f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2

，而

f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1

。只有最后一个为负。

因此，正确答案是 E。

The functional equation makes

f

completely additive: for

x = \prod p^{e_p},

we have

f(x) = \sum e_p\, f(p) = \sum e_p\, p,

where a prime in the denominator contributes a negative exponent (since

f(\frac{1}{p}) = -p

).

Evaluating:

f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7,

f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3,

f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1,

f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2,

and

f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1.

Only the last is negative.

Thus, the correct answer is E.

2021 AMC 12A Spring 第 18 题

答案：E

难度评级：1950

小提示：

大提示：

解答：

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