2020 AMC 12A Problems/Problem 13

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

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, $b$ must be a factor of 36.

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

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2020 AMC 12A Problems/Problem 13

Contents

Problem

Solution

Solution 2

See Also