Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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+ | ==Problem== | ||
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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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==Solution 1== | ==Solution 1== | ||
− | Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>. A perfect square must have even powers | + | Prime factorizing <math>201^9</math>, we get <math>3^9\cdot67^9</math>. |
+ | 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 <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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Observe that <math>201 = 67 \cdot 3</math>. Now divide into cases: | ||
+ | |||
+ | '''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>. | ||
+ | |||
+ | '''Case 2''': The factor is <math>67^n</math>. This is the same as Case 1. | ||
+ | |||
+ | '''Case 3''': The factor is some combination of <math>3</math>s and <math>67</math>s. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | <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. | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | <math>n = 4</math> is a "square". | ||
+ | |||
+ | <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. | ||
+ | |||
+ | <math>n = 8</math> is a "square" | ||
+ | |||
+ | <math>n = 9</math> is a "cube". | ||
+ | |||
+ | Now let's consider subcases: | ||
+ | |||
+ | '''Subcase 1''': The squares are with each other. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | '''Subcase 2''': The cubes are with each other. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | '''Subcase 3''': A number pairs with <math>n=6</math>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/XZiO19KNiYA | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/JR1LpMc3Ntg | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
− | + | {{AMC10 box|year=2019|ab=A|num-b=10|num-a=12}} | |
+ | {{MAA Notice}} |
Revision as of 22:57, 20 November 2020
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 ( , , , and ), we get .
Solution 2
Observe that . Now divide into cases:
Case 1: The factor is . Then we can have , , , , , or .
Case 2: The factor is . This is the same as Case 1.
Case 3: The factor is some combination of s and s.
This would be easy if we could just have any combination, as that would simply give . 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 .
is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of .
is a "square".
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.
is a "square"
is a "cube".
Now let's consider subcases:
Subcase 1: The squares are with each other.
Since we have square terms, and they would pair with other square terms, we get possibilities.
Subcase 2: The cubes are with each other.
Since we have cube terms, and they would pair with other cube terms, we get possibilities.
Subcase 3: A number pairs with .
Since any number can pair with (as it gives both a square and a cube), there would be possibilities. Remember however that there can be two different bases ( and ), and they would produce different results. Thus, there are in fact possibilities.
Finally, summing the cases gives .
Video Solution
Education, the Study of Everything
Video Solution
~savannahsolver
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.