2021 AMC 12A Problems/Problem 25

Revision as of 16:29, 11 February 2021 by Lopkiloinm (talk | contribs) (Solution)

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 (ProblemsAnswer KeyResources)
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