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Difference between revisions of "2021 AMC 12A Problems/Problem 25"

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==Solution==
 
==Solution==
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Start off with the number x that is 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>\frac{3}{\sqrt[3]{9}} > 1</math> because <math>\frac{3}{\sqrt[3]{9}}^3 > 1^3</math> because <math>3 < 1</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 is 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>\frac{3}{\sqrt[3]{9}} > 1</math> because <math>\frac{3}{\sqrt[3]{9}}^3 > 1^3</math> because <math>3 < 1</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 15:29, 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 is not divisible by 3. Multiply x by 9. Multiplying x by 9 adds a set $D(x)$ divisors that are the original divisors multiply by 3 and an additional $d(x)$ divisors that are the originals multiplied by 9 which end up saying that $d(9x) = 3d(x)$. Another consequence is multiplying the denominator by ${\sqrt[3]{9}}$. So now $f(9x) > f(x)$ because $\frac{d(9x)}{\sqrt[3]{9x}}= \frac{d(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}} > \frac{d(x)}{\sqrt[3]{x}}$ because $\frac{3}{\sqrt[3]{9}} > 1$ because $\frac{3}{\sqrt[3]{9}}^3 > 1^3$ because $3 < 1$. 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 (Problems • Answer Key • Resources)
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