2021 AMC 12A Problems/Problem 25
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[hide]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 every factor with a fixed
where
the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime
we look for the
for which
is a relative maximum:
Finally, the number we seek is The sum of its digits is
Actually, once we get that is a factor of
we know 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.
Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving )
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
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