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：

其他年份的第 18 题