Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"
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We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | ||
− | In this case, all <math>n</math> are 18, 50, 12, 20, 45, 28, 44. | + | In this case, all <math>n</math> are <math>18, 50, 12, 20, 45, 28, 44</math>. |
<math>\textbf{Case 9}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>. | <math>\textbf{Case 9}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>. | ||
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We have <math>f_1 \left( n \right) = 20</math>, <math>f_2 \left( n \right) = f_1 \left( 20 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | We have <math>f_1 \left( n \right) = 20</math>, <math>f_2 \left( n \right) = f_1 \left( 20 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | ||
− | In this case, the only <math>n</math> is 48. | + | In this case, the only <math>n</math> is <math>48</math>. |
<math>\textbf{Case 11}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^2 p_2^2</math>. | <math>\textbf{Case 11}</math>: the prime factorization of <math>n</math> takes the form <math>p_1^2 p_2^2</math>. | ||
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We have <math>f_1 \left( n \right) = 18</math>, <math>f_2 \left( n \right) = f_1 \left( 18 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | We have <math>f_1 \left( n \right) = 18</math>, <math>f_2 \left( n \right) = f_1 \left( 18 \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | ||
− | In this case, the only <math>n</math> is 36. | + | In this case, the only <math>n</math> is <math>36</math>. |
<math>\textbf{Case 12}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2 p_3</math>. | <math>\textbf{Case 12}</math>: the prime factorization of <math>n</math> takes the form <math>p_1 p_2 p_3</math>. |
Revision as of 18:20, 4 December 2021
- The following problem is from both the 2021 Fall AMC 10A #23 and 2021 Fall AMC 12A #20, so both problems redirect to this page.
Problem
For each positive integer , let be twice the number of positive integer divisors of , and for , let . For how many values of is
Solution 1
First, we can test values that would make true. For this to happen must have divisors, which means its prime factorization is in the form or , where and are prime numbers. Listing out values less than which have these prime factorizations, we find for , and just for . Here especially catches our eyes, as this means if one of , each of will all be . This is because (as given in the problem statement), so were , plugging this in we get , and thus the pattern repeats. Hence, as long as for a , such that and , must be true, which also immediately makes all our previously listed numbers, where , possible values of .
We also know that if were to be any of these numbers, would satisfy as well. Looking through each of the possibilities aside from , we see that could only possibly be equal to and , and still have less than or equal to . This would mean must have , or divisors, and testing out, we see that will then be of the form , or . The only two values less than or equal to would be and respectively. From here there are no more possible values, so tallying our possibilities we count values (Namely ).
~Ericsz
Solution 2
: .
Hence, if has the property that for some , then for all .
: .
Hence, if has the property that for some , then for all .
: .
We have , , , . Hence, Observation 2 implies .
: is prime.
We have , , . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have , . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have , . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have . Hence, Observation 1 implies .
In this case the only is .
: the prime factorization of takes the form .
We have . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have . Hence, Observation 1 implies .
In this case, all are .
: the prime factorization of takes the form .
We have , , . Hence, Observation 2 implies .
: the prime factorization of takes the form .
We have , . Hence, Observation 1 implies .
In this case, the only is .
: the prime factorization of takes the form .
We have , . Hence, Observation 1 implies .
In this case, the only is .
: the prime factorization of takes the form .
We have , , . Hence, Observation 2 implies .
Putting all cases together, the number of feasible is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=WQQVjCdoqWI
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.