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 makes it now <math>\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}</math>. <math>f(9x) > f(x)</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>
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Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 makes f now <math>\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}</math>. <math>f(9x) > f(x)</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>
 
~Lopkiloinm
 
~Lopkiloinm
  

Revision as of 15:15, 11 February 2021

Problem

Let $d(n)$ denote the number of positive integers that divide $n$, including $1$ and $n$. For example, $d(1)=1,d(2)=2,$ and $d(12)=6$. (This function is known as the divisor function.) Let\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$. What is the sum of the digits of $N?$

$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution

Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 makes f now $\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}$. $f(9x) > f(x)$. A property of multiples of 9 is their digits add up to multiples of 9, so the only possibility is $\boxed{(E) 9}$ ~Lopkiloinm

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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