Difference between revisions of "2020 AMC 12A Problems/Problem 13"
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<cmath>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</cmath> | <cmath>\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}</cmath> | ||
− | for all <math>N | + | 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> | ||
− | == Solution == | + | == 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>. | ||
− | + | 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 | + | 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 | + | 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 | ~Silverdragon | ||
− | Edits by ~Snore | + | Edits by ~Snore, ~Swaggergotcha |
== Solution 3 == | == 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. | 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>. | ||
− | + | ~jackshi2006 | |
==See Also== | ==See Also== |
Latest revision as of 18:48, 12 September 2023
Problem
There are integers and
each greater than
such that
for all . What is
?
Solution 1
can be simplified to
The equation is then which implies that
has to be
since
.
is the result when
and
are
and
being
will make the fraction
which is close to
.
Finally, with being
, the fraction becomes
. In this case
and
work, which means that
must equal
~lopkiloinm
Solution 2
As above, notice that you get
Now, combine the fractions to get .
Let us assume and
as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get . Note also that from the second equation,
and
must be factors of 36.
After listing out the factors of 36 and utilising trial and error, we find that and
works, with
. So our answer is
~Silverdragon
Edits by ~Snore, ~Swaggergotcha
Solution 3
Collapsed, . Comparing this to
, observe that
and
. The first can be rewritten as
. Then,
has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows
and
. Then,
, 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
can be rewritten as
. The trick is to focus on the properties of the denominators and numerators.
must be either be a factor of 36, because in any other situation the denominator of the final expression is preserved, and
is forced to be 36, which is not an answer choice. Next, realize that the numerator of
must be
, and even more importantly, that the denominator must be large enough so that when
is subtracted from it, it will form a fraction less than
. The only number large enough to do this will also being a fraction of
is 18. This means that
is 2. Moving onward, we are left with
. Since
must be a factor of
, just plug in values from the answer choices to find that
and
.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.