Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"
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− | ==Problem== | + | {{duplicate|[[2021 Fall AMC 10A Problems/Problem 23|2021 Fall AMC 10A #23]] and [[2021 Fall AMC 12A Problems/Problem 20|2021 Fall AMC 12A #20]]}} |
+ | |||
+ | == Problem == | ||
For each positive integer <math>n</math>, let <math>f_1(n)</math> be twice the number of positive integer divisors of <math>n</math>, and for <math>j \ge 2</math>, let <math>f_j(n) = f_1(f_{j-1}(n))</math>. For how many values of <math>n \le 50</math> is <math>f_{50}(n) = 12?</math> | For each positive integer <math>n</math>, let <math>f_1(n)</math> be twice the number of positive integer divisors of <math>n</math>, and for <math>j \ge 2</math>, let <math>f_j(n) = f_1(f_{j-1}(n))</math>. For how many values of <math>n \le 50</math> is <math>f_{50}(n) = 12?</math> | ||
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==Solution== | ==Solution== | ||
+ | First, we can test values that would make <math>f(x)=12</math> true. For this to happen <math>x</math> must have <math>6</math> divisors, which means its prime factorization is in the form <math>pq^2</math> or <math>p^5</math>, where <math>p</math> and <math>q</math> are prime numbers. Listing out values less than <math>50</math> which have these prime factorizations, we find <math>12,20,28,44,18,45,50</math> for <math>pq^2</math>, and just <math>32</math> for <math>p^5</math>. Here <math>12</math> especially catches our eyes, as this means if one of <math>f_i(n)=12</math>, each of <math>f_{i+1}(n), f_{i+2}(n), ...</math> will all be <math>12</math>. This is because <math>f_{i+1}(n)=f(f_i(n))</math> (as given in the problem statement), so were <math>f_i(n)=12</math>, plugging this in we get <math>f_{i+1}(n)=f(12)=12</math>, and thus the pattern repeats. Hence, as long as for a <math>i</math>, such that <math>i\leq 50</math> and <math>f_{i}(n)=12</math>, <math>f_{50}(n)=12</math> must be true, which also immediately makes all our previously listed numbers, where <math>f(x)=12</math>, possible values of <math>n</math>. | ||
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+ | We also know that if <math>f(x)</math> were to be any of these numbers, <math>x</math> would satisfy <math>f_{50}(n)</math> as well. Looking through each of the possibilities aside from <math>12</math>, we see that <math>f(x)</math> could only possibly be equal to <math>20</math> and <math>18</math>, and still have <math>x</math> less than or equal to <math>50</math>. This would mean <math>x</math> must have <math>10</math>, or <math>9</math> divisors, and testing out, we see that <math>x</math> will then be of the form <math>p^4q</math>, or <math>p^2q^2</math>. The only two values less than or equal to <math>50</math> would be <math>48</math> and <math>36</math> respectively. From here there are no more possible values, so tallying our possibilities we count<math>\boxed{D:10}</math> values (Namely <math>12,20,28,44,18,45,50,32,36,48</math>). | ||
+ | |||
+ | ~Ericsz | ||
+ | |||
+ | ==Video Solution by Mathematical Dexterity== | ||
+ | https://www.youtube.com/watch?v=WQQVjCdoqWI | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021 Fall|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2021 Fall|ab=A|num-b=19|num-a=21}} | ||
+ | {{AMC10 box|year=2021 Fall|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:21, 23 November 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
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
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.