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

Revision as of 21:20, 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

Solution 3

Collapsed, $$ (Error compiling LaTeX. Unknown error_msg)\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1} $. Comparing this to$ \sqrt[36]{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{D)}3$, as only 4 and 3 factor into 36 and 24 while being 1 apart.

(b=1 technically works but I don't care. a,b,c>1 as in the question)

~~BJHHar

@@ Line 34: / Line 34: @@
 ~Silverdragon
+== Solution 3 ==
+Collapsed, <math></math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}<math>. Comparing this to </math>\sqrt[36]{N^{25}}<math>, observe that </math>bc+c+1=25<math> and </math>abc=36<math>. The first can be rewritten as </math>c(b+1)=24<math>. Then, </math>b+1<math> has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows </math>36=2^2 3^2<math> and </math>24=2^33<math>. Then, </math>b=\boxed{\textbf{D)}3$, as only 4 and 3 factor into 36 and 24 while being 1 apart.
+(b=1 technically works but I don't care. a,b,c>1 as in the question)
+~~BJHHar
 ==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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