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

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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
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==See Also==
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{{AMC12 box|year=2020|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 15:15, 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

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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

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