# 2020 AMC 12A Problems/Problem 13

## Problem

There are integers $a, b,$ and $c,$ each greater than $1,$ such that $$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt{N^{25}}$$

for all $N \neq 1$. 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 $\frac{bc+c+1}{abc}=\frac{25}{36}$.

WLOG, let $bc+c+1=25$ 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 trial and error we find that $c=6$ and $b=3$ works, with $a=2$. So our answer is $\boxed{\textbf{(B) } 3.}$

~Silverdragon

Edits by ~Snore

## Solution 3

Collapsed, $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}}$. Comparing this to $\sqrt{N^{25}}$, observe that $bc+c+1=25$ and $abc=36$. The first can be rewritten as $c(b+1)=24$. Then, $b+1$ has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows $36=2^2 3^2$ and $24=2^33$. Then, $b=\boxed{\textbf{B)}3}$, as only 4 and 3 factor into 36 and 24 while being 1 apart.

~~BJHHar

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 