Question

有多少个有理数四元组 $(a, b, c, d)$ 满足 $\begin{aligned} &a\log_{10} 2 + b\log_{10} 3 \\ &\quad {}+ c\log_{10} 5 + d\log_{10} 7 = 2005? \end{aligned}$

How many distinct four-tuples $(a, b, c, d)$ of rational numbers are there with $\begin{aligned} &a\log_{10} 2 + b\log_{10} 3 \\ &\quad {}+ c\log_{10} 5 + d\log_{10} 7 = 2005? \end{aligned}$

$0$

$1$

$17$

$2004$

无限多个

infinitely many

答案：B

知识点：对数质因数分解

难度评级：1800

解答：

原方程等价于 $\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005$ ，所以 $2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}.$

把 $a, b, c, d$ 的分母乘以同一个整数消去，再用质因数分解的唯一性比较指数，可得 $a = 2005$ 、 $b = 0$ 、 $c = 2005$ 、 $d = 0$ 。

所以恰有 $1$ 个这样的四元组。

所以正确答案是 B。

The equation is equivalent to $\log_{10}\left(2^a 3^b 5^c 7^d\right) = 2005,$ so $2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005} \cdot 5^{2005}.$

Clearing the denominators of $a, b, c, d$ with a common integer multiplier and using the uniqueness of prime factorization, the exponents must match: $a = 2005,$ $b = 0,$ $c = 2005,$ and $d = 0.$

So there is exactly $1$ such four-tuple.

Thus, the correct answer is B.

2005 AMC 12B 第 17 题

答案：B

难度评级：1800

解答：

其他年份的第 17 题