AMC12 2025 A

AMC12 2025 A · Q7

AMC12 2025 A · Q7. It mainly tests Exponents & radicals, Logarithms (rare).

In a certain alien world, the maximum running speed $v$ of an organism is dependent on its number of toes $n$ and number of eyes $m$. The relationship can be expressed as $v = kn^am^b$ centimeters per hour, where k, a, b are integer constants. In a population where all organisms have 5 toes, $\log v = 4+2\log m$; and in a population where all organisms have 25 eyes, $\log v = 4 + 4 \log n$, where all logs are in base 10. What is $k+a+b$?

在某个外星世界，有机体的最大奔跑速度$v$取决于其脚趾数$n$和眼睛数$m$。关系可表示为$v = kn^am^b$厘米/小时，其中$k, a, b$为整数常数。在所有有机体都有5个脚趾的种群中，$\log v = 4+2\log m$；在所有有机体都有25只眼睛的种群中，$\log v = 4 + 4 \log n$，所有对数均为10为底。求$k+a+b$？

(A) 20 20

(B) 21 21

(D) 23 23

(E) 24 24

Answer

Correct choice: (C)

正确答案：（C）

Solution

Substituting $v$ in the equation where $n=5$, we have: $\log(k \cdot 5^a m^b)$ $=$ $4+2\log m$. Using logarithmic properties, we can write this as: $\log(k \cdot 5^a m^b)$ $=$ $\log(10^4 \cdot m^2)$ We can do this with the other equation where m=25: $\log(k \cdot n^a 25^b)$ $=$ $\log(10^4 \cdot n^4)$ Now we can get rid of the logs on both sides and are left with the following system of equations: $k \cdot 5^am^b = 10^4 m^2$ $k \cdot n^a25^b = 10^4 n^4$ Notice that in the first equation, we can change $m$ arbitrarily, so we know that the exponent of $m$ must be the same - hence $b=2$. Similarly, from the second equation, we get $a=4$. $10^4$ can be written as $2^4 \cdot 5^4$, which means that $k=2^4 = 16$. Thus the answer is $2+4+16 = \boxed{(C) 22}$.

将$n=5$代入方程，我们有： $\log(k \cdot 5^a m^b)$ $=$ $4+2\log m$。使用对数性质，可写成： $\log(k \cdot 5^a m^b)$ $=$ $\log(10^4 \cdot m^2)$ 对$m=25$的另一个方程同理： $\log(k \cdot n^a 25^b)$ $=$ $\log(10^4 \cdot n^4)$ 现在去掉两边对数，得到方程组： $k \cdot 5^am^b = 10^4 m^2$ $k \cdot n^a25^b = 10^4 n^4$ 注意第一个方程中$m$可任意变化，因此$m$的指数必须相同，故$b=2$。类似地，从第二个方程得$a=4$。$10^4$可写成$2^4 \cdot 5^4$，因此$k=2^4 = 16$。于是答案是$16+4+2 = \boxed{(C) 22}$。

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