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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>
 
<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==
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==Solution 1 (PIE)==
 
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 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.  
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A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get 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.  
  
 
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>.  
  
 
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>.
 
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>.
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~ Continuous_Pi
  
 
==Solution 2==
 
==Solution 2==
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Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>.
 
Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>.
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== Solution 3 (A Little Long) ==
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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.
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<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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~ PerseverePlayer
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 00:16, 3 September 2025

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 a 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 $41-4 = \boxed{\textbf{(C) }37}$.

~ Continuous_Pi

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 (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)
Preceded by
Problem 10 		Followed by
Problem 12
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 10 Problems and Solutions

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