2020 AMC 12A Problems/Problem 13: Difference between revisions

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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

@@ Line 27: / Line 27: @@
 Now, combine the fractions to get <math>\frac{bc+c+1}{abc}=\frac{25}{36}</math>.
-Let us assume <math>bc+c+1=25</math> and <math>abc=36</math> as this is most convenient. (EDIT: This used to say WLOG but that is inaccurate)
+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 both be factors of 36.
+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 some 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>
+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
+Edits by ~Snore, ~Swaggergotcha
 == Solution 3 ==

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
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