2021 Fall AMC 12A Problems/Problem 20
- 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.