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

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There are integers <math>a, b,</math> and <math>c,</math> each greater than <math>1,</math> such that
 
There are integers <math>a, b,</math> and <math>c,</math> each greater than <math>1,</math> such that
  
−
<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</math>
+
<cmath>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</cmath>
 +
 
 +
for all <math>N > 1</math>. What is <math>b</math>?
  
 
<math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6</math>
 
<math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6</math>

Revision as of 06:39, 2 February 2020

Problem

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

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

for all $N > 1$. What is $b$?

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

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

Solution 2

As above, notice that you get $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$

Now, combine the fractions to get $\frac{bc+c+1}{abc}=\frac{25}{36}$.

Assume that $bc+c+1=25$ and $abc=36$.

From the first equation we get $c(b+1)=24$. Note also that from the second equation, $b$ and $c$ must both be factors of 36.

After some casework we find that $c=6$ and $b=3$ works, with $a=2$. So our answer is $\boxed{\textbf{(B) } 3.}$

~Silverdragon

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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