Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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Note that the values <math>\frac{2}{\sqrt[3]{3}}</math> and <math>\frac{3}{\sqrt[3]{9}}</math> do not have to be explicitly computed; we only need the fact that <math>\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1</math> which is easy to show by hand. | Note that the values <math>\frac{2}{\sqrt[3]{3}}</math> and <math>\frac{3}{\sqrt[3]{9}}</math> do not have to be explicitly computed; we only need the fact that <math>\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1</math> which is easy to show by hand. | ||
− | The above claim automatically implies <math>N</math> is a multiple of 9: if <math>N</math> was not divisible by 9, then <math>f(9N) > f(N)</math> which is a contradiction, and if <math>N</math> was divisible by 3 and not 9, then <math>f(3N) > f(N) > f\left(\frac{N}{3}\right)</math>, also a contradiction. Then the sum of digits of <math>N</math> must be a multiple of 9, so only choice <math>\textbf{(E) } 9</math> works. | + | The above claim automatically implies <math>N</math> is a multiple of 9: if <math>N</math> was not divisible by 9, then <math>f(9N) > f(N)</math> which is a contradiction, and if <math>N</math> was divisible by 3 and not 9, then <math>f(3N) > f(N) > f\left(\frac{N}{3}\right)</math>, also a contradiction. Then the sum of digits of <math>N</math> must be a multiple of 9, so only choice <math>\boxed{\textbf{(E) } 9}</math> works. |
== Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) == | == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) == |
Revision as of 23:51, 13 February 2021
Contents
Problem
Let denote the number of positive integers that divide , including and . For example, and . (This function is known as the divisor function.) LetThere 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) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.