Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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− | https://youtu.be/6P-0ZHAaC_A (by OmegaLearn) | + | https://youtu.be/6P-0ZHAaC_A (by OmegaLearn) ~ pi_is_3.14 |
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==See also== | ==See also== |
Revision as of 16:10, 20 June 2021
Problem
Let denote the number of positive integers that divide
, including
and
. For example,
and
. (This function is known as the divisor function.) Let
There is a unique positive integer
such that
for all positive integers
. What is the sum of the digits of
Solution 1
Consider the prime factorization By the Multiplication Principle,
Now, we rewrite
as
As
for all positive integers
it follows that for all positive integers
and
if and only if
So,
is maximized if and only if
is maximized.
For each independent factor with a fixed
where
the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. For each prime
with
we look for the
for which
is a relative maximum:
Finally, the positive integer we seek is The sum of its digits is
Alternatively, once we notice that is a factor of
we can conclude that the sum of the digits of
must be a multiple of
Only choice
is possible.
~MRENTHUSIASM
Solution 2 (Fast)
Using the answer choices to our advantage, we can show that must be divisible by 9 without explicitly computing
, by exploiting the following fact:
Claim: If is not divisible by 3, then
.
Proof: Since is a multiplicative function, we have
and
. Then
Note that the values
and
do not have to be explicitly computed; we only need the fact that
which is easy to show by hand.
The above claim automatically implies is a multiple of 9: if
was not divisible by 9, then
which is a contradiction, and if
was divisible by 3 and not 9, then
, also a contradiction. Then the sum of digits of
must be a multiple of 9, so only choice
works.
-scrabbler94
Video Solutions
https://www.youtube.com/watch?v=gWaUNz0gLE0 (by Dedekind Cuts)
https://www.youtube.com/watch?v=Sv4gj1vMjOs (by Aaron He)
https://youtube.com/watch?v=y_7s8fvMCdI (by Punxsutawney Phil)
https://youtu.be/6P-0ZHAaC_A (by OmegaLearn) ~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last problem |
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 |
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