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

Difference between revisions of "2020 AMC 12A Problems/Problem 13"

(→‎Solution)
(→‎Solution)
Line 13: Line 13:
 
The equation is then <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}</math> which implies that <math>\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.</math>
 
The equation is then <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}</math> which implies that <math>\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.</math>
  
−
<math>a</math> has to be <math>2</math> since <math>\frac{25}{36}>\frac{1}{2}</math>.  
+
<math>a</math> has to be <math>2</math> since <math>\frac{25}{36}>\frac{7}{12}</math>. <math>\frac{7}{12}</math> is the result when <math>a, b,</math> and <math>c</math> are <math>3, 2,</math> and <math>2</math> 
  
−
<math>b</math> being <math>3</math> will make the fraction <math>frac{2}{3}</math> which is close to <math>frac{25}{36}</math>.  
+
<math>b</math> being <math>3</math> will make the fraction <math>\frac{2}{3}</math> which is close to <math>\frac{25}{36}</math>.  
  
−
Finally, with <math>c</math> being <math>6</math>, the fraction becomes <math>frac{25}{36}</math>. In this case <math>a, b,</math> and <math>c</math> work, which means that <math>b</math> must equal <math>\boxed{\textbf{(B) } 3.}</math>~lopkiloinm
+
Finally, with <math>c</math> being <math>6</math>, the fraction becomes <math>\frac{25}{36}</math>. In this case <math>a, b,</math> and <math>c</math> work, which means that <math>b</math> must equal <math>\boxed{\textbf{(B) } 3.}</math>~lopkiloinm

Revision as of 16:12, 1 February 2020

Problem

There are integers $a, b,$ and $c,$ each greater than $1,$ such that

$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}$

$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution

$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$

The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$

$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$. $\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$

$b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$.

Finally, with $c$ being $6$, the fraction becomes $\frac{25}{36}$. In this case $a, b,$ and $c$ work, which means that $b$ must equal $\boxed{\textbf{(B) } 3.}$~lopkiloinm

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_13&oldid=116322"