Question

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 $f(N) \gt f(n)$ for all positive integers $n \ne N.$ What is the sum of the digits of $N?$

$5$

$6$

$7$

$8$

$9$

Solution:

Since $f(n) = \dfrac{d(n)}{n^{1/3}}$ is multiplicative, its value factors over prime powers as a product of terms $\dfrac{e + 1}{p^{e/3}}$ for each prime power $p^e\ \|\ n.$ We maximize each term separately.

For $p = 2,$ the ratio $\dfrac{e+1}{2^{e/3}}$ is largest at $e = 3$ (value $2$ ). For $p = 3,$ it peaks at $e = 2;$ for $p = 5$ and $p = 7,$ at $e = 1;$ and for every prime $p \ge 11,$ the best choice is $e = 0$ (the ratio already drops below $1$ at $e = 1$ ).

Hence $N = 2^3\cdot 3^2\cdot 5\cdot 7 = 2520,$ whose digit sum is $2 + 5 + 2 + 0 = 9.$

Thus, the correct answer is E.

2021 AMC 12A Spring Problem 25

Solution:

Problem 25 in Other Years