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| − | == Problem ==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_20]] |
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| − | 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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| − | <math>\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11</math>
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| − | ==Solution==
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| − | 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>).
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| − | ~Ericsz
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| − | ==Video Solution by Mathematical Dexterity==
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| − | https://www.youtube.com/watch?v=WQQVjCdoqWI
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=22|num-a=24}}
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| − | {{MAA Notice}}
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