2021 AMC 12A Problems/Problem 25

Revision as of 19:37, 11 February 2021 by Pi is 3.14 (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

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


Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving )

https://youtu.be/6P-0ZHAaC_A

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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