2021 AMC 12A Problems/Problem 25

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

Suppose a counting number x be 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 $3 > \sqrt[3]{9}$ . A property of multiples of 9 is their digits add up to multiples of 9, so the only possibility is $\boxed{(E) 9}$

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

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2021 AMC 12A Problems/Problem 25

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