Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

← Older edit
Cyrus825 (talk | contribs)
editor
35 edits
No edit summary
Piggydd (talk | contribs)
editor
7 edits
 
(50 intermediate revisions by 8 users not shown)
Line 1: Line 1:
Solution 1:
==Problem==
Taking the logarithm of both sides and using n = 5, 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>?


<imath>log</imath> <imath>k</imath> * <imath>5^am^b</imath> <imath>=</imath> <imath>4+2logm</imath>. We can now use logarithmic properties to rewrite this as:
<imath>\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24</imath>
<imath>log</imath> <imath>k</imath> * <imath>5^am^b</imath> <imath>=</imath> <imath>log10^4*m^2</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}}

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

See Also

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2025_AMC_12A_Problems/Problem_7&oldid=268214"