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

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==Solution 1==
 
==Solution 1==
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Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>. A perfect square must have even powers from 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. Perfect cubes must have multiples of 3 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 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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Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>.  
 +
A perfect square must have even powers from 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.  
 +
 
 +
Perfect cubes must have multiples of 3 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 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>.
  
 
Solution by Aadileo
 
Solution by Aadileo

Revision as of 22:34, 9 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 from 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 six ($3^0\cdot67^0$ , $3^0\cdot67^6$ , $3^6\cdot67^0$, and $3^6\cdot67^6$), we get $41-4 = \boxed{\textbf{(C) }37}$.

Solution by Aadileo

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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