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Difference between revisions of "2021 AMC 12A Problems/Problem 25"

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Start off with the number 1. Multiply 1 by 2. When you multiply 1 by 2, you double the number of divisors and you divide it by <math>\sqrt[3]{2}</math>. Multiply 2 by 2. When you multiply 2 by 2, you triple and then halve the number of divisors. You must not forget to divide it by <math>\sqrt[3]{2}</math> again to get that function. Keep up multiplying by 2 until it starts to decrease. Keep on multiplying by 3 until it starts to decrease. Keep on multiplying by 5 until it starts to decrease. Keep on multiplying by 7 until it starts to decrease. You cannot multiply by 11 because first multiplication automatically causes a decrease. <math>1 \rightarrow 2 \rightarrow 2^2 \rightarrow 2^3 \rightarrow 2^33 \rightarrow 2^32^2 \rightarrow 2^33^25 \rightarrow2^33^257</math> This is like the computer science algorithms.
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Start off with the number x that does not have a factor of 3. Multiply x by 9. Multiplying x by 9 triples the number of divisors but also leads to a decrease in the function because of the denominator being multiplied by <math>\sqrt[3]{9}</math>. The number is now <math>\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}</math> If I am not mistaken, multiplying a number by 9 leads to an increase of a property of multiples of 9 is their digits add up to 9, so the only possibility is <math>\boxed{(D) 9}</math>
 
~Lopkiloinm
 
~Lopkiloinm
  

Revision as of 15:06, 11 February 2021

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 does not have a factor of 3. Multiply x by 9. Multiplying x by 9 triples the number of divisors but also leads to a decrease in the function because of the denominator being multiplied by $\sqrt[3]{9}$. The number is now $\frac{D(x)}{\sqrt[3]{x}}\frac{3}{\sqrt[3]{9}}$ If I am not mistaken, multiplying a number by 9 leads to an increase of a property of multiples of 9 is their digits add up to 9, so the only possibility is $\boxed{(D) 9}$ ~Lopkiloinm

Note

See problem 1.

See also

2021 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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