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Difference between revisions of "2019 AMC 10A Problems/Problem 11"

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How many positive integer divisors of <math>201^9</math> are perfect squares or perfect cubes (or both)?
 
How many positive integer divisors of <math>201^9</math> are perfect squares or perfect cubes (or both)?
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<math>\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41</math>
  
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<math>{\textbf{(A) }32} \qquad {\textbf{(B) }36} \qquad {\textbf{(C) }37} \qquad {\textbf{(D) }39} \qquad {\textbf{(E) }41}</math>
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==Solution 1 (PIE)==
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==Solution 1==
 
 
Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>.  
 
Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>.  
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A perfect square must have even powers of its prime factors, so our possible choices for our exponents of a perfect square are <math>0, 2, 4, 6, 8</math> for both <math>3</math> and <math>67</math>. This yields <math>5\cdot5 = 25</math> perfect squares.  
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A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are <math>0, 2, 4, 6, 8</math> for both <math>3</math> and <math>67</math>. This yields <math>5\cdot5 = 25</math> perfect squares.  
  
 
Perfect cubes must have multiples of <math>3</math> for each of their prime factors' exponents, so we have either <math>0, 3, 6</math>, or <math>9</math> for both <math>3</math> and <math>67</math>, which yields <math>4\cdot4 = 16</math> perfect cubes, for a total of <math>25+16 = 41</math>.  
 
Perfect cubes must have multiples of <math>3</math> for each of their prime factors' exponents, so we have either <math>0, 3, 6</math>, or <math>9</math> for both <math>3</math> and <math>67</math>, which yields <math>4\cdot4 = 16</math> perfect cubes, for a total of <math>25+16 = 41</math>.  
  
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Subtracting the overcounted powers of six (<math>3^0\cdot67^0</math> , <math>3^0\cdot67^6</math> , <math>3^6\cdot67^0</math>, and <math>3^6\cdot67^6</math>), we get <math>41-4 = \boxed{\textbf{(C) }37}</math>.
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Subtracting the overcounted powers of <math>6</math> (<math>3^0\cdot67^0</math> , <math>3^0\cdot67^6</math> , <math>3^6\cdot67^0</math>, and <math>3^6\cdot67^6</math>), we get <math>41-4 = \boxed{\textbf{(C) }37}</math>.
 +
 
 +
==Solution 2==
 +
Observe that <math>201 = 67 \cdot 3</math>. Now divide into cases:
 +
 
 +
'''Case 1''': The factor is <math>3^n</math>. Then we can have <math>n = 2</math>, <math>3</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>9</math>.  
  
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Solution by Aadileo
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'''Case 2''': The factor is <math>67^n</math>. This is the same as Case 1.
  
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==Solution 2==
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'''Case 3''': The factor is some combination of <math>3</math>s and <math>67</math>s.
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201 = 67 * 3. Split the answers into cases:
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 +
This would be easy if we could just have any combination, as that would simply give <math>6 \cdot 6</math>. However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for <math>n</math>.
 +
 
 +
<math>n = 2</math> is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
 +
 
 +
<math>n = 3</math> is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of <math>3</math>.
 +
 
 +
<math>n = 4</math> is a "square".
 +
 
 +
<math>n = 6</math> is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.
 +
 
 +
<math>n = 8</math> is a "square"
 +
 
 +
<math>n = 9</math> is a "cube".
 +
 
 +
Now let's consider subcases:
 +
 
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'''Subcase 1''': The squares are with each other.
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Since we have <math>3</math> square terms, and they would pair with <math>3</math> other square terms, we get <math>3 \cdot 3 = 9</math> possibilities.
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'''Subcase 2''': The cubes are with each other.
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 +
Since we have <math>2</math> cube terms, and they would pair with <math>2</math> other cube terms, we get <math>2 \cdot 2 = 4</math> possibilities.
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'''Subcase 3''': A number pairs with <math>n=6</math>.
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Since any number can pair with <math>n=6</math> (as it gives both a square and a cube), there would be <math>6</math> possibilities. Remember however that there can be two different bases (<math>3</math> and <math>67</math>), and they would produce different results. Thus, there are in fact <math>6 \cdot 2 = 12</math> possibilities.
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Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>.
 +
 
 +
== Solution 3 (Quick) ==
 +
 
 +
We first prime factorize <math>201^9 = 3^9 \cdot 67^9</math>. Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of <math>3</math> in the exponent. The largest square for <math>3</math> can be <math>3^8</math>, so their must be <math>\dfrac {8}{2} = 4</math> ways. The largest cube is <math>3^9</math>, so there must be <math>\dfrac{9}{3} = 3</math>. Minus one <math>3^6</math> due to overlapping and we get <math>4 + 3 -1 = 6</math> ways for <math>3</math> to be a cube/square. We can see that this same thing happens for <math>67^9</math> due to the same exponent. Adding <math>0</math> as a case, we have our answer; <math>6 \cdot 6 + 1 = \boxed{\textbf{(C) }37}</math>
 +
 
 +
~ Wiselion
 +
 
 +
== Solution 4 (A Little Long) ==
 +
 
 +
Notice that <math>201=3 \cdot 67</math>. We factorize <math>201^9</math> to get <math>3^9 \cdot 67^9</math>. We then list perfect squares and cubes.
 +
<math>3^2</math>, <math>3^4</math>, <math>3^6</math>, <math>3^8</math>. <math>3^3</math>, <math>3^6</math>, <math>3^9</math>. <math>67^2</math>, <math>67^4</math>, <math>67^6</math>, <math>67^8</math>. <math>67^3</math>, <math>67^6</math>, <math>67^9</math>. Notice that the powers of <math>6</math> overlap. We must not forget <math>1</math> though. Of course, all of these factors already work. This gives us <math>15-2=3</math>. Next, we count the perfect squares. Since there are <math>4</math> options we have <math>4 \cdot 4=16</math>. We do the same for the perfect cubes except with 3 options this time, and we have <math>3 \cdot 3=9</math>. However, we accidentally overcounted <math>3^6 \cdot 67^6</math>. We add our answers and subtract <math>1</math> to get <math>13+16+9-1 = \boxed{\textbf{(C) }37}</math>
  
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Case 1: The factor is 3^n
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~ PerseverePlayer
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Then, we would have n = 2, 3, 4, 6, 8, and 9.
 
  
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Case 2: The factor is 67^n
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==Video Solution==
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Same as case 1
+
https://youtu.be/JR1LpMc3Ntg
  
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Case 3: The factor is some combination
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~savannahsolver
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This would be easy if we could just have any combination, as that would be simply 6*6. However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our n values.
 
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n = 2 is a "square" because it would be a factor of this number that is a perfect square. More generally, it is even.
 
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n = 3 is a "cube" because it would be a factor of this number that is a perfect cube. More generally, it is a multiple of 3.
 
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n = 4 is a "square"
 
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n = 6 is interesting, since its both a "square" and a "cube". Don't count this as either because it's double function would double count the numbers if we counted them simply as a "square" or as a "cube", so we will count them in another case.
 
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n = 8 is a "square"
 
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n = 9 is a "cube".
 
  
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Now let's do subcases:
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== Video Solution by OmegaLearn==
 +
https://youtu.be/ZhAZ1oPe5Ds?t=2402
  
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Subcase 1: The squares are with each other.
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~ pi_is_3.14
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Since we have 3 square terms, and they would pair with 3 other square terms, we get 3*3 = 9 cases
 
  
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Subcase 2: The cubes are with each other.
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==Video Solution==
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Since we have 2 cube terms, and they would pair with 2 other cube terms, we get 2*2 = 4 cases
 
  
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Subcase 3: A number pairs with n=6.
+
https://youtu.be/XZiO19KNiYA
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Since any number can pair with n=6 since it's a square AND cube, there are 6 cases. Remember however that there can be two different bases (3 and 67), and they would produce different results. Thus, there are 6*2 = 12 cases
 
  
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Thus, we add up the cases, to get 6+6+9+4+12 = 37.
+
Education, the Study of Everything
  
−
iron
 
  
 
==See Also==
 
==See Also==

Latest revision as of 13:05, 26 November 2023

Problem

How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$

Solution 1 (PIE)

Prime factorizing $201^9$, we get $3^9\cdot67^9$. A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$. This yields $5\cdot5 = 25$ perfect squares.

Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0, 3, 6$, or $9$ for both $3$ and $67$, which yields $4\cdot4 = 16$ perfect cubes, for a total of $25+16 = 41$.

Subtracting the overcounted powers of $6$ ($3^0\cdot67^0$ , $3^0\cdot67^6$ , $3^6\cdot67^0$, and $3^6\cdot67^6$), we get $41-4 = \boxed{\textbf{(C) }37}$.

Solution 2

Observe that $201 = 67 \cdot 3$. Now divide into cases:

Case 1: The factor is $3^n$. Then we can have $n = 2$, $3$, $4$, $6$, $8$, or $9$.

Case 2: The factor is $67^n$. This is the same as Case 1.

Case 3: The factor is some combination of $3$s and $67$s.

This would be easy if we could just have any combination, as that would simply give $6 \cdot 6$. However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for $n$.

$n = 2$ is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.

$n = 3$ is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of $3$.

$n = 4$ is a "square".

$n = 6$ is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.

$n = 8$ is a "square"

$n = 9$ is a "cube".

Now let's consider subcases:

Subcase 1: The squares are with each other.

Since we have $3$ square terms, and they would pair with $3$ other square terms, we get $3 \cdot 3 = 9$ possibilities.

Subcase 2: The cubes are with each other.

Since we have $2$ cube terms, and they would pair with $2$ other cube terms, we get $2 \cdot 2 = 4$ possibilities.

Subcase 3: A number pairs with $n=6$.

Since any number can pair with $n=6$ (as it gives both a square and a cube), there would be $6$ possibilities. Remember however that there can be two different bases ($3$ and $67$), and they would produce different results. Thus, there are in fact $6 \cdot 2 = 12$ possibilities.

Finally, summing the cases gives $6+6+9+4+12 = \boxed{\textbf{(C) }37}$.

Solution 3 (Quick)

We first prime factorize $201^9 = 3^9 \cdot 67^9$. Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of $3$ in the exponent. The largest square for $3$ can be $3^8$, so their must be $\dfrac {8}{2} = 4$ ways. The largest cube is $3^9$, so there must be $\dfrac{9}{3} = 3$. Minus one $3^6$ due to overlapping and we get $4 + 3 -1 = 6$ ways for $3$ to be a cube/square. We can see that this same thing happens for $67^9$ due to the same exponent. Adding $0$ as a case, we have our answer; $6 \cdot 6 + 1 = \boxed{\textbf{(C) }37}$

~ Wiselion

Solution 4 (A Little Long)

Notice that $201=3 \cdot 67$. We factorize $201^9$ to get $3^9 \cdot 67^9$. We then list perfect squares and cubes. $3^2$, $3^4$, $3^6$, $3^8$. $3^3$, $3^6$, $3^9$. $67^2$, $67^4$, $67^6$, $67^8$. $67^3$, $67^6$, $67^9$. Notice that the powers of $6$ overlap. We must not forget $1$ though. Of course, all of these factors already work. This gives us $15-2=3$. Next, we count the perfect squares. Since there are $4$ options we have $4 \cdot 4=16$. We do the same for the perfect cubes except with 3 options this time, and we have $3 \cdot 3=9$. However, we accidentally overcounted $3^6 \cdot 67^6$. We add our answers and subtract $1$ to get $13+16+9-1 = \boxed{\textbf{(C) }37}$

~ PerseverePlayer

Video Solution

https://youtu.be/JR1LpMc3Ntg

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=2402

~ pi_is_3.14

Video Solution

https://youtu.be/XZiO19KNiYA

Education, the Study of Everything


See Also

2019 AMC 10A (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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