AMC12 2005 A

AMC12 2005 A · Q21

AMC12 2005 A · Q21. It mainly tests Exponents & radicals, Manipulating equations.

How many ordered triples of integers $(a,b,c)$, with $a \geq 2$, $b \geq 1$, and $c \geq 0$, satisfy both $\log_{a}b = c^{2005}$ and $a + b + c = 2005$?

有整数的有序三元组 $(a,b,c)$ 多少个，其中 $a \geq 2$，$b \geq 1$，$c \geq 0$，满足 $\log_{a}b = c^{2005}$ 和 $a + b + c = 2005$？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (C)

正确答案：（C）

Solution

$a^{c^{2005}} = b$ Casework upon $c$: - $c = 0$: Then $a^0 = b \Longrightarrow b = 1$. Thus we get $(2004,1,0)$. - $c = 1$: Then $a^1 = b \Longrightarrow a = b$. Thus we get $(1002,1002,1)$. - $c \ge 2$: Then the exponent of $a$ becomes huge, and since $a \ge 2$ there is no way we can satisfy the second condition. Hence we have two ordered triples $\mathrm{(C)}$.

$a^{c^{2005}} = b$ 对 $c$ 分情况讨论： - $c = 0$：则 $a^0 = b \Longrightarrow b = 1$。因此得到 $(2004,1,0)$。 - $c = 1$：则 $a^1 = b \Longrightarrow a = b$。因此得到 $(1002,1002,1)$。 - $c \ge 2$：则 $a$ 的指数会变得极大，并且由于 $a \ge 2$，不可能满足第二个条件。因此共有两个有序三元组 $\mathrm{(C)}$。

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