Difference between revisions of "2019 AMC 10A Problems/Problem 11"
Ironicninja (talk | contribs) (→Solution 2) |
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Case 1: The factor is 3^n | Case 1: The factor is 3^n | ||
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Then, we would have n = 2, 3, 4, 6, 8, and 9. | Then, we would have n = 2, 3, 4, 6, 8, and 9. | ||
Case 2: The factor is 67^n | Case 2: The factor is 67^n | ||
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Same as case 1 | Same as case 1 | ||
Case 3: The factor is some combination | Case 3: The factor is some combination | ||
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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. | 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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n = 2 is a "square" because it would be a factor of this number that is a perfect square. More generally, it is even. | 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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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. | 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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n = 4 is a "square" | n = 4 is a "square" | ||
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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. | 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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n = 8 is a "square" | n = 8 is a "square" | ||
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n = 9 is a "cube". | n = 9 is a "cube". | ||
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Subcase 1: The squares are with each other. | Subcase 1: The squares are with each other. | ||
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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 | Since we have 3 square terms, and they would pair with 3 other square terms, we get 3*3 = 9 cases | ||
Subcase 2: The cubes are with each other. | Subcase 2: The cubes are with each other. | ||
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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 | Since we have 2 cube terms, and they would pair with 2 other cube terms, we get 2*2 = 4 cases | ||
Subcase 3: A number pairs with n=6. | Subcase 3: A number pairs with n=6. | ||
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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 | 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 | ||
Revision as of 20:14, 11 February 2019
Contents
[hide]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
Solution 2
201 = 67 * 3. Split the answers into cases:
Case 1: The factor is 3^n
Then, we would have n = 2, 3, 4, 6, 8, and 9.
Case 2: The factor is 67^n
Same as case 1
Case 3: The factor is some combination
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.
n = 2 is a "square" because it would be a factor of this number that is a perfect square. More generally, it is even.
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.
n = 4 is a "square"
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.
n = 8 is a "square"
n = 9 is a "cube".
Now let's do 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*3 = 9 cases
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*2 = 4 cases
Subcase 3: A number pairs with n=6.
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
Thus, we add up the cases, to get 6+6+9+4+12 = 37.
iron
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.