Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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==Solution== | ==Solution== | ||
| − | Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 triples the number of divisors but also leads to a decrease in the function because of the denominator being multiplied by <math>\sqrt[3]{9}</math>. The number is now <math>\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}</math> | + | Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 triples the number of divisors but also leads to a decrease in the function because of the denominator being multiplied by <math>\sqrt[3]{9}</math>. The number is now <math>\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}</math>. Multiplying a non multiple of 3 by 9 Multiplying a number by 9 leads to an increase of a property of multiples of 9 is their digits add up to 9, so the only possibility is <math>\boxed{(E) 9}</math> |
~Lopkiloinm | ~Lopkiloinm | ||
Revision as of 15:07, 11 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.) Let![\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]](http://latex.artofproblemsolving.com/4/1/e/41e0a31875f03c4b386a006e694adbb03555a964.png)
There is a unique positive integer 
such that 
for all positive integers 
. What is the sum of the digits of 
Solution
Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 triples the number of divisors but also leads to a decrease in the function because of the denominator being multiplied by ![$\sqrt[3]{9}$](http://latex.artofproblemsolving.com/a/7/3/a736f2264b72e806f66318ff1bbbc64d5053d964.png)
. The number is now ![$\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}$](http://latex.artofproblemsolving.com/a/e/c/aec4682310f3fd76a640cb538e023538740c39d7.png)
. Multiplying a non multiple of 3 by 9 Multiplying a number by 9 leads to an increase of a property of multiples of 9 is their digits add up to 9, so the only possibility is 
~Lopkiloinm
Note
See problem 1.
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. 
