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2020 AMC 12A Problems/Problem 13: Difference between revisions

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<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}</math> can be simplified to <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.</math>
<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}</math> can be simplified to <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.</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>a<math> has to be </math>2<math> since </math>\frac{25}{36}>\frac{1}{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>. 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.}$~lopkiloinm
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>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

Revision as of 15:07, 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 $\frac{25}{36}>\frac{1}{2}$.

$b$ 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

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