2025 AMC 12A Problems/Problem 7

Problem

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 $k+a+b$ ?

$\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24$

Solution 1 (logs and system of equations)

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 $2+4+16 = \boxed{(C) 22}$ .

-Cyrus825 ~ScoutViolet (mostly minor fixes) ~knight10 (minor fixes)

Solution 2

We first try to simplify both log equations, and then we bring in the equation for velocity.

For the equation representing organisms with 5 toes:

$\log v = 4 + 2 \log m$

$\log v = \log 10^4 + \log m^2$

$\log v = \log(10^4\cdot m^2)$

$v = 10^4\cdot m^2$

$k\cdot 5^am^b = 10^4\cdot m^2$

We do the same with the logarithm equation representing organisms with 25 eyes to get:

$\log v = 4 + 4 \log n$

$v = 10^4\cdot n^4$

$k\cdot n^a(25^b) = 10^4\cdot n^4$

Now we need to figure out what $a$ , $b$ , and $k$ are. Looking at the first equation (or the second), we notice that we can easily match factors. There are two $m$ s on the right side so we set $b = 2$ so there are two $m$ s on the left side. Now we need $10^4 = k\cdot 5^a = 5^4\cdot 2^4$ . There are four $5$ s on one side, so we need $a = 4$ to get four $5$ s on the other. Now, $k = 2^4 = 16$ . This solution works for the first equation.

Checking if these values work with the second equation, it does, so our answer is $2 + 4 + 16 = \boxed{22}$

~Logibyte

Video Solution by Power Solve

https://youtu.be/Vd_kvodRjNQ?si=V7ea9tJxYVGf9Uz7&t=420

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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