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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>
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<cmath>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</cmath>
 +
 
 +
for all <math>N \neq 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>
  
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== Solution ==
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== Solution 1==
  
 
<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}</math> can be simplified to <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.</math>
 
<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}</math> can be simplified to <math>N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.</math>
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Now, combine the fractions to get <math>\frac{bc+c+1}{abc}=\frac{25}{36}</math>.
 
Now, combine the fractions to get <math>\frac{bc+c+1}{abc}=\frac{25}{36}</math>.
  
−
Assume that <math>bc+c+1=25</math> and <math>abc=36</math>.
+
Let us assume <math>bc+c+1=25</math> and <math>abc=36</math> as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
 +
 
 +
From the first equation, we get <math>c(b+1)=24</math>. Note also that from the second equation, <math>b</math> and <math>c</math> must be factors of 36.
 +
 
 +
After listing out the factors of 36 and utilising trial and error, we find that <math>c=6</math> and <math>b=3</math> works, with <math>a=2</math>. So our answer is <math>\boxed{\textbf{(B) } 3.}</math>
 +
 
 +
~Silverdragon
 +
 
 +
Edits by ~Snore, ~Swaggergotcha
 +
 
 +
== Solution 3 ==
 +
Collapsed, <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{B)}3}</math>, as only 4 and 3 factor into 36 and 24 while being 1 apart.
 +
 
 +
~~BJHHar
 +
 
 +
Note: 3 and 2 also factor into 24 and 36 while being 1 apart, but we can eliminate by quick guess and check. ~ Edit by Soupnoodle
 +
 
 +
== Solution 4==
 +
<math>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</math> can be rewritten as <math>\frac{\frac{\frac{1}{c}+1}{b}+1}{a} = \frac{25}{36}</math>. The trick is to focus on the properties of the denominators and numerators. <math>a</math> must be either be a factor of 36, because in any other situation the denominator of the final expression is preserved, and <math>b</math> is forced to be 36, which is not an answer choice. Next, realize that the numerator of <math>\frac{\frac{1}{c}+1}{b}+1</math> must be <math>25</math>, and even more importantly, that the denominator must be large enough so that when <math>1</math> is subtracted from it, it will form a fraction less than <math>1</math>. The only number large enough to do this will also being a fraction of <math>36</math> is 18. This means that <math>a</math> is 2. Moving onward, we are left with <math>\frac{\frac{1}{c}+1}{b} = \frac{7}{18}</math>. Since <math>b</math> must be a factor of <math>18</math>, just plug in values from the answer choices to find that <math>c=6</math> and <math>b=3</math>.
  
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From the first equation we get <math>c(b+1)=24</math>. Note also that from the second equation, <math>b</math> must be a factor of 36.
+
~jackshi2006
  
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After some casework we find that <math>c=6</math> and <math>b=3</math> works, with <math>a=2</math>. So our answer is <math>\boxed{\textbf{(B) } 3.}</math>
 
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2020|ab=A|num-b=12|num-a=14}}
 
{{AMC12 box|year=2020|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:48, 12 September 2023

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 $N \neq 1$. What is $b$?

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

Solution 1

$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ 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 $\frac{bc+c+1}{abc}=\frac{25}{36}$.

Let us assume $bc+c+1=25$ and $abc=36$ as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)

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

After listing out the factors of 36 and utilising 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, ~Swaggergotcha

Solution 3

Collapsed, $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}}$. 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, $b=\boxed{\textbf{B)}3}$, as only 4 and 3 factor into 36 and 24 while being 1 apart.

~~BJHHar

Note: 3 and 2 also factor into 24 and 36 while being 1 apart, but we can eliminate by quick guess and check. ~ Edit by Soupnoodle

Solution 4

$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}$ can be rewritten as $\frac{\frac{\frac{1}{c}+1}{b}+1}{a} = \frac{25}{36}$. The trick is to focus on the properties of the denominators and numerators. $a$ must be either be a factor of 36, because in any other situation the denominator of the final expression is preserved, and $b$ is forced to be 36, which is not an answer choice. Next, realize that the numerator of $\frac{\frac{1}{c}+1}{b}+1$ must be $25$, and even more importantly, that the denominator must be large enough so that when $1$ is subtracted from it, it will form a fraction less than $1$. The only number large enough to do this will also being a fraction of $36$ is 18. This means that $a$ is 2. Moving onward, we are left with $\frac{\frac{1}{c}+1}{b} = \frac{7}{18}$. Since $b$ must be a factor of $18$, just plug in values from the answer choices to find that $c=6$ and $b=3$.

~jackshi2006

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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