Difference between revisions of "2021 Fall AMC 12A Problems/Problem 20"
m |
m (→Solution 2) |
||
Line 56: | Line 56: | ||
We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | We have <math>f_1 \left( n \right) = 12</math>. Hence, Observation 1 implies <math>f_{50} \left( n \right) = 12</math>. | ||
− | In this case, all <math>n</math> are <math>18, | + | In this case, all <math>n</math> are <math>12, 18, 20, 28, 44, 45,</math> and <math>50</math>. |
<math>\textbf{Case 9}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>. | <math>\textbf{Case 9}</math>: The prime factorization of <math>n</math> takes the form <math>p_1 p_2^3</math>. |
Revision as of 03:04, 6 June 2022
- 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
:
.
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.