Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"

(Created page with "==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>,...")
 
Line 1: Line 1:
==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>
  
Line 5: Line 7:
  
 
==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>.
 +
 +
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 $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution

First, we can test values that would make $f(x)=12$ true. For this to happen $x$ must have $6$ divisors, which means its prime factorization is in the form $pq^2$ or $p^5$, where $p$ and $q$ are prime numbers. Listing out values less than $50$ which have these prime factorizations, we find $12,20,28,44,18,45,50$ for $pq^2$, and just $32$ for $p^5$. Here $12$ especially catches our eyes, as this means if one of $f_i(n)=12$, each of $f_{i+1}(n), f_{i+2}(n), ...$ will all be $12$. This is because $f_{i+1}(n)=f(f_i(n))$ (as given in the problem statement), so were $f_i(n)=12$, plugging this in we get $f_{i+1}(n)=f(12)=12$, and thus the pattern repeats. Hence, as long as for a $i$, such that $i\leq 50$ and $f_{i}(n)=12$, $f_{50}(n)=12$ must be true, which also immediately makes all our previously listed numbers, where $f(x)=12$, possible values of $n$.

We also know that if $f(x)$ were to be any of these numbers, $x$ would satisfy $f_{50}(n)$ as well. Looking through each of the possibilities aside from $12$, we see that $f(x)$ could only possibly be equal to $20$ and $18$, and still have $x$ less than or equal to $50$. This would mean $x$ must have $10$, or $9$ divisors, and testing out, we see that $x$ will then be of the form $p^4q$, or $p^2q^2$. The only two values less than or equal to $50$ would be $48$ and $36$ respectively. From here there are no more possible values, so tallying our possibilities we count$\boxed{D:10}$ values (Namely $12,20,28,44,18,45,50,32,36,48$).

~Ericsz

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=WQQVjCdoqWI

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png