2025 AMC 12A Problems/Problem 7: Difference between revisions

Latest revision as of 22:56, 9 November 2025

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

@@ Line 1: / Line 1: @@
-Solution 1:
+==Problem==
-Taking the logarithm of both sides in the first equation, we have:
+In a certain alien world, the maximum running speed <imath>v</imath> of an organism is dependent on its number of toes <imath>n</imath> and number of eyes <imath>m</imath>. The relationship can be expressed as <imath>v = kn^am^b</imath> centimeters per hour, where k, a, b are integer constants. In a population where all organisms have 5 toes, <imath>\log v = 4+2\log m</imath>; and in a population where all organisms have 25 eyes, <imath>\log v = 4 + 4 \log n</imath>, where all logs are in base 10. What is <imath>k+a+b</imath>?
-log v = log k<imath>n^a</imath> <imath>m^b</imath>
+<imath>\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24</imath>
+== Solution 1 (logs and system of equations) ==
+Substituting <imath>v</imath> in the equation where <imath>n=5</imath>, we have:
+<imath>\log(k \cdot 5^a m^b)</imath> <imath>=</imath> <imath>4+2\log m</imath>.
+Using logarithmic properties, we can write this as:
+<imath>\log(k \cdot 5^a m^b)</imath> <imath>=</imath> <imath>\log(10^4 \cdot m^2)</imath>
+We can do this with the other equation where m=25:
+<imath>\log(k \cdot n^a 25^b)</imath> <imath>=</imath> <imath>\log(10^4 \cdot n^4)</imath>
+Now we can get rid of the logs on both sides and are left with the following system of equations:
+<imath>k \cdot 5^am^b = 10^4 m^2</imath>
+<imath>k \cdot n^a25^b = 10^4 n^4</imath>
+Notice that in the first equation, we can change <imath>m</imath> arbitrarily, so we know that the exponent of <imath>m</imath> must be the same - hence <imath>b=2</imath>. Similarly, from the second equation, we get <imath>a=4</imath>. <imath>10^4</imath> can be written as <imath>2^4 \cdot 5^4</imath>, which means that <imath>k=2^4 = 16</imath>. Thus the answer is <imath>2+4+16 = \boxed{(C)  22}</imath>.
+-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:
+<imath>\log v = 4 + 2 \log m</imath>
+<imath>\log v = \log 10^4 + \log m^2</imath>
+<imath>\log v = \log(10^4\cdot m^2)</imath>
+<imath>v = 10^4\cdot m^2</imath>
+<imath>k\cdot 5^am^b = 10^4\cdot m^2</imath>
+We do the same with the logarithm equation representing organisms with 25 eyes to get:
+<imath>\log v = 4 + 4 \log n</imath>
+<imath>v = 10^4\cdot n^4</imath>
+<imath>k\cdot n^a(25^b) = 10^4\cdot n^4</imath>
+Now we need to figure out what <imath>a</imath>, <imath>b</imath>, and <imath>k</imath> are. Looking at the first equation (or the second), we notice that we can easily match factors. There are two <imath>m</imath>s on the right side so we set <imath>b = 2</imath> so there are two <imath>m</imath>s on the left side. Now we need <imath>10^4 = k\cdot 5^a = 5^4\cdot 2^4</imath>. There are four <imath>5</imath>s on one side, so we need <imath>a = 4</imath> to get four <imath>5</imath>s on the other. Now, <imath>k = 2^4 = 16</imath>. This solution works for the first equation.
+Checking if these values work with the second equation, it does, so our answer is <imath>2 + 4 + 16 = \boxed{22}</imath>
+~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
+==See Also==
+{{AMC12 box|year=2025|ab=A|num-b=6|num-a=8}}
+* [[AMC 12]]
+* [[AMC 12 Problems and Solutions]]
+* [[Mathematics competitions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

2025 AMC 12A (Problems • Answer Key • Resources)
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