Difference between revisions of "2021 AMC 12A Problems/Problem 25"
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Suppose a counting number x be not divisible by 3. Multiply x by 9. Multiplying x by 9 adds a set <math>d(x)</math> divisors that are the original divisors multiply by 3 and an additional <math>d(x)</math> divisors that are the originals multiplied by 9 which end up saying that <math>d(9x) = 3d(x)</math>. Another consequence is multiplying the denominator by <math>{\sqrt[3]{9}}</math>. So now <math>f(9x) > f(x)</math> because <math>\frac{d(9x)}{\sqrt[3]{9x}}= \frac{d(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}} > \frac{d(x)}{\sqrt[3]{x}}</math> because <math>3 > \sqrt[3]{9}</math>. A property of multiples of 9 is their digits add up to multiples of 9, so the only possibility is <math>\boxed{(E) 9}</math> | Suppose a counting number x be not divisible by 3. Multiply x by 9. Multiplying x by 9 adds a set <math>d(x)</math> divisors that are the original divisors multiply by 3 and an additional <math>d(x)</math> divisors that are the originals multiplied by 9 which end up saying that <math>d(9x) = 3d(x)</math>. Another consequence is multiplying the denominator by <math>{\sqrt[3]{9}}</math>. So now <math>f(9x) > f(x)</math> because <math>\frac{d(9x)}{\sqrt[3]{9x}}= \frac{d(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}} > \frac{d(x)}{\sqrt[3]{x}}</math> because <math>3 > \sqrt[3]{9}</math>. A property of multiples of 9 is their digits add up to multiples of 9, so the only possibility is <math>\boxed{(E) 9}</math> | ||
| − | Edit: It seems that this proof is not complete because we also need to check whether multiply by 3 is better than multiply by 9. It is better to multiply by 9 than by 3 shown by similar logic | + | Edit: It seems that this proof is not complete because we also need to check whether multiply by 3 is better than multiply by 9. It is better to multiply by 9 than by 3 shown by similar logic which I will leave as an exercise for the reader. |
~Lopkiloinm | ~Lopkiloinm | ||
Revision as of 17:56, 11 February 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![\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]](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
Suppose a counting number x be not divisible by 3. Multiply x by 9. Multiplying x by 9 adds a set 
divisors that are the original divisors multiply by 3 and an additional divisors that are the originals multiplied by 9 which end up saying that 
. Another consequence is multiplying the denominator by ![${\sqrt[3]{9}}$](http://latex.artofproblemsolving.com/a/e/9/ae9f43b95da16ed95fef1833151e1e950fb0b07e.png)
. So now 
because ![$\frac{d(9x)}{\sqrt[3]{9x}}= \frac{d(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}} > \frac{d(x)}{\sqrt[3]{x}}$](http://latex.artofproblemsolving.com/b/4/4/b4467ad166eace0ee3126bcc0fde76a9a1c8d7d4.png)
because ![$3 > \sqrt[3]{9}$](http://latex.artofproblemsolving.com/7/8/e/78ee79490e3266311df7f1f343202c37ae38fb0d.png)
. A property of multiples of 9 is their digits add up to multiples of 9, so the only possibility is 
Edit: It seems that this proof is not complete because we also need to check whether multiply by 3 is better than multiply by 9. It is better to multiply by 9 than by 3 shown by similar logic which I will leave as an exercise for the reader. ~Lopkiloinm
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. 
