Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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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== | ||
| + | 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. In total, we have <math>25+16 = 41</math>. However, we have overcounted perfect 6'ths: <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 must subtract these 4, for our final answer, which is <math>41-4 = \boxed{\textbf{(C) }37}</math>. | ||
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| + | Solution by Aadileo | ||
Revision as of 16:37, 9 February 2019
How many positive integer divisors of 
are perfect squares or perfect cubes (or both)?
Solution 1
Prime factorizing , we get 
. A perfect square must have even powers from its prime factors, so our possible choices for our exponents of a perfect square are 
for both 
and 
. This yields 
perfect squares. Perfect cubes must have multiples of 3 for each of their prime factors' exponents, so we have either 
, or 
for both and , which yields 
perfect cubes. In total, we have 
. However, we have overcounted perfect 6'ths: 
, 
, 
, and 
. We must subtract these 4, for our final answer, which is 
.
Solution by Aadileo
