2005 AMC 12B 第 17 题

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}\text{？}

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

把左边改写为

\log_{10}\left(2^a 3^b 5^c 7^d\right)

。

Rewrite the left side as

\log_{10}\left(2^a 3^b 5^c 7^d\right)

于是

2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005}5^{2005}

；用唯一分解匹配指数。

Then

2^a 3^b 5^c 7^d = 10^{2005} = 2^{2005}5^{2005}

; match exponents via unique factorization

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}\text{。}

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.

其他年份的第 17 题