# 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 1 (Generalization)

We consider the prime factorization of $n:$ $$n=\prod_{i=1}^{k}p_i^{e_i}.$$ By the Multiplication Principle, we have $$d(n)=\prod_{i=1}^{k}(e_i+1).$$ Now, we rewrite $f(n)$ as $$f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.$$ As $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if $$f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}$$ is maximized.

For each independent factor $\frac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ we look for the nonnegative integer $e_i$ such that $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: $$\begin{array}{c|c|c|c|c} & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex] \hline\hline & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\ & & 1 & 4 & \\ & & 2 & 27/4 &\\ & & 3 & 8 & \checkmark\\ & & 4 & 125/16 & \\ [0.5ex] \hline & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\ & & 1 & 8/3 & \\ & & 2 & 3 & \checkmark\\ & & 3 & 64/27 & \\ [0.5ex] \hline & & & & \\ [-2ex] 3 & 5 & 0 & 1 & \\ & & 1 & 8/5 & \checkmark\\ & & 2 & 27/25 & \\ [0.5ex] \hline & & & & \\ [-2ex] 4 & 7 & 0 & 1 & \\ & & 1 & 8/7 & \checkmark\\ & & 2 & 27/49 & \\ [0.5ex] \hline & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\ & & \geq1 & \leq8/11 & \\ [0.5ex] \end{array}$$ Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{\textbf{(E) }9}.$

Alternatively, once we notice that $3^2$ is a factor of $N,$ we can conclude that the sum of the digits of $N$ must be a multiple of $9.$ Only choice $\textbf{(E)}$ is possible.

~MRENTHUSIASM

## Solution 2 (Solution 1 but Fewer Notations)

The question statement asks for the value of $N$ that maximizes $f(N)$. Let $N$ start out at $1$; we will find what factors to multiply $N$ by, in order for $N$ to maximize the function.

First, we will find what power of $2$ to multiply $N$ by. If we multiply $N$ by $2^{a}$, the numerator of $f$, $d(N)$, will multiply by a factor of $a+1$; this is because the number $2^{a}$ has $a+1$ divisors. The denominator, $\sqrt[3]{N}$, will simply multiply by $\sqrt[3]{2^{a}}$. Therefore, the entire function multiplies by a factor of $\frac{a+1}{\sqrt[3]{2^{a}}}$. We want to find the integer value of $a$ that maximizes this value. By inspection, this is $3$. Therefore, we multiply $N$ by $8$; right now, $N$ is $8$.

Next, we will find what power of $3$ to multiply $N$ by. Similar to the previous step, we wish to find the integer value of $a$ that maximizes $\frac{a+1}{\sqrt[3]{3^{a}}}$. This value, also by inspection, is $2$.

We can repeat this step on the rest of the primes to get $$N = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7$$ but from $11$ on, $a=0$ will maximize the value of the function, so the prime is not a factor in $N$. We evaluate $N$ to be $2520$, so the answer is $\boxed{\textbf{(E) }9}$.

## Solution 3 (Fast)

Using the answer choices to our advantage, we can show that $N$ must be divisible by 9 without explicitly computing $N$, by exploiting the following fact:

Claim: If $n$ is not divisible by 3, then $f(9n) > f(3n) > f(n)$.

Proof: Since $d(\cdot)$ is a multiplicative function, we have $d(3n) = d(3)d(n) = 2d(n)$ and $d(9n) = 3d(n)$. Then \begin{align*} f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\ f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n) \end{align*} Note that the values $\frac{2}{\sqrt[3]{3}}$ and $\frac{3}{\sqrt[3]{9}}$ do not have to be explicitly computed; we only need the fact that $\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1$ which is easy to show by hand.

The above claim automatically implies $N$ is a multiple of 9: if $N$ was not divisible by 9, then $f(9N) > f(N)$ which is a contradiction, and if $N$ was divisible by 3 and not 9, then $f(3N) > f(N) > f\left(\frac{N}{3}\right)$, also a contradiction. Then the sum of digits of $N$ must be a multiple of 9, so only choice $\boxed{\textbf{(E) } 9}$ works.

-scrabbler94