Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"
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− | == Solution 2 == | + | == Solution 2 (*Rigorous* reasoning on why there cannot be any other solutions) == |
+ | First, take note that the maximum possible value of <math>f_1(n)</math> for <math>1 \le n \le k</math> increases as <math>k</math> increases (it is a step function), i.e. it is increasing. Likewise, as <math>k</math> decreases, the maximum possible value of <math>f_1(n)</math> decreases as well. Also, let <math>f_1(n) = 2d(n)</math> where <math>d(n)</math> is the number of divisors of n. | ||
+ | |||
+ | Since <math>n \le 50</math>, <math>f_1(n) <= 20</math>. This maximum occurs when <math>d(n) = 10 \implies n = 2^4 \cdot 3 = 48</math>. Next, since <math>f_1(n) <=20</math>, <math>f_1(f_1(n)) \le 12 \implies f_2(n) \le 12</math>. This maximum occurs when <math>d(f_1(n)) = 6 \implies n = 2 \cdot 3^2 = 18, n = 2^2 \cdot 3 = 12</math>. Since <math>f_2(n) \le 12</math>, <math>f_1(f_2(n)) \le 12 \implies f_3(n) \le 12</math>, once again. This maximum again occurs when <math>d(f_2(n)) = 6 \implies f_2(n) = 2^2 \cdot 3 = 12</math>. Now, suppose for the sake of contradiction that <math>f_2(n) < 12</math>. Then, <math>f_3(n) < 12</math> (since <math>f_2(n) = 12</math> was the only number that would maximize <math>f_3(n))</math> for <math>f_2(n) \le 12</math>). As a result, since <math>f_1(n)</math> is increasing, and because <math>12</math> is where <math>f_1</math> steps down from a maximum of <math>6 \cdot 2 = 12</math>, we must have that <math>f_1(f_3(n)) < f_1(12) = 12 \implies f_4(n) < 12</math>. We continue applying <math>f_1</math> on both sides (which is possible since <math>f_1</math> is increasing) until we reach <math>f_50</math>, giving us that <math>f_50(n) < 12</math>. However, <math>f_50(n) = 12</math>, which is a contradiction. Thus, <math>f_2(n) = 12</math>. | ||
+ | |||
+ | Now, let us finally solve for the solutions. <math>f_2(n) = 12 \implies f_1(f_1(n)) = 12 \implies d(f_1(n)) = 6</math>. <math>d(f_1(n)) = 6 \implies f_1(n) = p^2 \cdot q</math> where <math>p</math> and <math>q</math> are primes. Since <math>f_1(n) \le 20</math>, <math>f_1(n)</math> can only be <math>12</math>, <math>18</math>, or <math>20</math>. If <math>f_1(n) = 12</math>, then <math>d(n) = 6 \implies n = p^5, p^2 \cdot q \implies n \in \{ 12, 18, 20, 28, 32, 44, 45, 50 \}</math>, resulting in 8 solutions. If <math>f_1(n) = 18</math>, then <math>d(n) = 9 \implies n = p^8, p^2 \cdot q^2 \implies n = 36</math>, giving us one more solution. Finally, <math>f_1(n) = 20 \implies d(n) = 10 \implies n = p^9, p^4 \cdot q \implies n = 48</math>. Thus, in total, we have <math>\boxed{\textbf{(D)} 10}</math> solutions. | ||
+ | == Solution 3 == | ||
<math>\textbf{Observation 1}</math>: <math>f_1 \left( 12 \right) = 12</math>. | <math>\textbf{Observation 1}</math>: <math>f_1 \left( 12 \right) = 12</math>. | ||
Revision as of 15:05, 25 November 2023
- 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.
Contents
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 (*Rigorous* reasoning on why there cannot be any other solutions)
First, take note that the maximum possible value of for increases as increases (it is a step function), i.e. it is increasing. Likewise, as decreases, the maximum possible value of decreases as well. Also, let where is the number of divisors of n.
Since , . This maximum occurs when . Next, since , . This maximum occurs when . Since , , once again. This maximum again occurs when . Now, suppose for the sake of contradiction that . Then, (since was the only number that would maximize for ). As a result, since is increasing, and because is where steps down from a maximum of , we must have that . We continue applying on both sides (which is possible since is increasing) until we reach , giving us that . However, , which is a contradiction. Thus, .
Now, let us finally solve for the solutions. . where and are primes. Since , can only be , , or . If , then , resulting in 8 solutions. If , then , giving us one more solution. Finally, . Thus, in total, we have solutions.
Solution 3
: .
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 and .
: 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
Video Solution by TheBeautyofMath
~IceMatrix
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.