2021 Fall AMC 10A Problems/Problem 23
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 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
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| All AMC 10 Problems and Solutions | ||
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