Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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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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<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> | ||
==Solution 1 (PIE)== | ==Solution 1 (PIE)== | ||
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 of its prime factors, so our possible choices for our exponents to get 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. | + | A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get 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>. | 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>. | 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>. | ||
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+ | ~ Continuous_Pi | ||
==Solution 2== | ==Solution 2== | ||
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== Solution 3 (Quick) == | == Solution 3 (Quick) == | ||
− | We first prime factorize <math>201^9 = 3^9 \cdot 67^9</math>. Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of <math>3</math> in the exponent. The largest square for <math>3</math> can be <math>3^8</math>, so | + | We first prime factorize <math>201^9 = 3^9 \cdot 67^9</math>. Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of <math>3</math> in the exponent. The largest square for <math>3</math> can be <math>3^8</math>, so there must be <math>\dfrac {8}{2} = 4</math> ways. The largest cube is <math>3^9</math>, so there must be <math>\dfrac{9}{3} = 3</math>. Minus one <math>3^6</math> due to overlapping and we get <math>4 + 3 -1 = 6</math> ways for <math>3</math> to be a cube/square. We can see that this same thing happens for <math>67^9</math> due to the same exponent. Adding <math>0</math> as a case, we have our answer; <math>6 \cdot 6 + 1 = \boxed{\textbf{(C) }37}</math> |
− | ~ Wiselion | + | ~Wiselion |
− | == Solution 4 == | + | == Solution 4 (A Little Long) == |
− | Notice that <math>201=3 \cdot 67</math>. We factorize <math>201^9</math> to get <math>3^9 \cdot 67^9</math>. We then list perfect squares and cubes | + | Notice that <math>201=3 \cdot 67</math>. We factorize <math>201^9</math> to get <math>3^9 \cdot 67^9</math>. We then list perfect squares and cubes. |
+ | <math>3^2</math>, <math>3^4</math>, <math>3^6</math>, <math>3^8</math>. <math>3^3</math>, <math>3^6</math>, <math>3^9</math>. <math>67^2</math>, <math>67^4</math>, <math>67^6</math>, <math>67^8</math>. <math>67^3</math>, <math>67^6</math>, <math>67^9</math>. Notice that the powers of <math>6</math> overlap. We must not forget <math>1</math> though. Of course, all of these factors already work. This gives us <math>15-2=3</math>. Next, we count the perfect squares. Since there are <math>4</math> options we have <math>4 \cdot 4=16</math>. We do the same for the perfect cubes except with 3 options this time, and we have <math>3 \cdot 3=9</math>. However, we accidentally overcounted <math>3^6 \cdot 67^6</math>. We add our answers and subtract <math>1</math> to get <math>13+16+9-1 = \boxed{\textbf{(C) }37}</math> | ||
~ PerseverePlayer | ~ PerseverePlayer |
Latest revision as of 18:25, 21 October 2024
Contents
[hide]Problem
How many positive integer divisors of are perfect squares or perfect cubes (or both)?
Solution 1 (PIE)
Prime factorizing , we get . A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get 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 .
~ Continuous_Pi
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 .
Solution 3 (Quick)
We first prime factorize . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of in the exponent. The largest square for can be , so there must be ways. The largest cube is , so there must be . Minus one due to overlapping and we get ways for to be a cube/square. We can see that this same thing happens for due to the same exponent. Adding as a case, we have our answer;
~Wiselion
Solution 4 (A Little Long)
Notice that . We factorize to get . We then list perfect squares and cubes. , , , . , , . , , , . , , . Notice that the powers of overlap. We must not forget though. Of course, all of these factors already work. This gives us . Next, we count the perfect squares. Since there are options we have . We do the same for the perfect cubes except with 3 options this time, and we have . However, we accidentally overcounted . We add our answers and subtract to get
~ PerseverePlayer
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=2402
~ pi_is_3.14
Video Solution
Education, the Study of Everything
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.