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

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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>.
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Solution by Aadileo
 
  
 
==Solution 2==
 
==Solution 2==
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201 = 67 * 3. Split the answers into cases:
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Observe that <math>201 = 67 \cdot 3</math>. Now divide into cases:
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Case 1: The factor is 3^n
 
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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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Same as case 1
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'''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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Case 3: The factor is some combination
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'''Case 2''': The factor is <math>67^n</math>. This is the same as Case 1.
  
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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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'''Case 3''': The factor is some combination of <math>3</math>s and <math>67</math>s.
  
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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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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>.
  
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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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<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.
  
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n = 4 is a "square"
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<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>.
  
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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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<math>n = 4</math> is a "square".
  
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n = 8 is a "square"
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<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.
  
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n = 9 is a "cube".
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<math>n = 8</math> is a "square"
  
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Now let's do subcases:
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<math>n = 9</math> is a "cube".
  
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Subcase 1: The squares are with each other.
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Now let's consider subcases:
  
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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 1''': The squares are with each other.
  
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Subcase 2: The cubes 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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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 2''': The cubes are with each other.
  
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Subcase 3: A number pairs with n=6.
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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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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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'''Subcase 3''': A number pairs with <math>n=6</math>.
  
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Thus, we add up the cases, to get 6+6+9+4+12 = 37.
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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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iron
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Finally, summing the cases gives <math>6+6+9+4+12 = 37</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:17, 17 February 2019

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

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

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 $6+6+9+4+12 = 37$.

See Also

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