Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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==Solution 1== | ==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>. | ||
| − | A perfect square must have even powers | + | 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. |
| − | 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>. | + | 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 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>. | 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>. | ||
Revision as of 19:41, 10 February 2019
Problem
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 of 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 for each of their prime factors' exponents, so we have either 
, or 
for both and , which yields 
perfect cubes, for a total of 
.
Subtracting the overcounted powers of six (
, 
, 
, and 
), we get 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
