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

(→‎Solution 1)
(combine solutions 2 and 3 with more rigorous justification why we can say N is a multiple of 9)
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2==
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==Solution 2 (Fast)==
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A cube root seems bad, so we should just cube it. It seems that if the number is a multiple of 3, there are only two choices. If the number is a multiple of 9, there is one choice. We can prove that for all k is indivisible by 3, f(9k) > f(3k) > f(k). The divisors of 3k contain the divisors of k and the divisors of k multiplied by 3. The divisors of 9k contain the divisors of k, the divisors of k multiplied by 3, and the divisors of k multiplied by 9. <cmath>\frac{27}{9}d(k)^3 > \frac{8}{3}d(k)^3 > d(k)^3</cmath> so <math>f(9k) > f(3k) > f(k)</math> and since <math>\boxed{\textbf{(E) }9}</math> is the only possible answer choice, it is the answer.
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Using the answer choices to our advantage, we can show that <math>N</math> must be divisible by 9 without explicitly computing <math>N</math>, by exploiting the following fact:
  
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==Solution 3 (Only Use If Low On Time)==
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'''Claim''': If <math>n</math> is not divisible by 3, then <math>f(9n) > f(3n) > f(n)</math>.
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We can guess that <math>N</math> would be divisible by <math>9</math>. Recall, for a number to be divisible by <math>9</math>, the sum of its digits must also be divisible by <math>9</math>. Since there's only one choice where it's divisible by <math>9</math>, we get <math>\boxed{\textbf{(E)} ~9}</math> as our answer. ~rocketsri
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'''Proof''': Since <math>d(\cdot)</math> is a [[multiplicative function]], we have <math>d(3n) = d(3)d(n) = 2d(n)</math> and <math>d(9n) = 3d(n)</math>. Then
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<cmath>\begin{align*}
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f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\
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f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n)
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\end{align*}
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</cmath>
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Note that the values <math>\frac{2}{\sqrt[3]{3}}</math> and <math>\frac{3}{\sqrt[3]{9}}</math> do not have to be explicitly computed; we only need the fact that <math>\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1</math> which is easy to show by hand.
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The above claim automatically implies <math>N</math> is a multiple of 9: if <math>N</math> was not divisible by 9, then <math>f(9N) > f(N)</math> which is a contradiction, and if <math>N</math> was divisible by 3 and not 9, then <math>f(3N) > f(N) > f\left(\frac{N}{3}\right)</math>, also a contradiction. Then the sum of digits of <math>N</math> must be a multiple of 9, so only choice <math>\textbf{(E) } 9</math> works.
  
 
== Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==
 
== Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==

Revision as of 22:48, 13 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 1

Consider the prime factorization \[n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.\] By the Multiplication Principle, \[d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).\] Now, we rewrite $f(n)$ as \[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).\] As $f(n)>0$ for all positive integers $n,$ it follows that for all positive integers $a$ and $b$, $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3.$ So, $f(n)$ is maximized if and only if \[f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)\] is maximized.

For every factor $\frac{(e_i+1)^3}{{p_i}^{e_i}}$ with a fixed $p_i$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime $p_i=2,3,5,7,\cdots,$ we look for the $e_i$ for which $\frac{(e_i+1)^3}{{p_i}^{e_i}}$ is a relative maximum: \[\begin{array}{ c c c c } p_i & e_i & (e_i+1)^3/\left({p_i}^{e_i}\right) & \text{max?} \\ \hline\hline 2 & 0 & 1 & \\ 2 & 1 & 4 & \\ 2 & 2 & 27/4 &\\ 2 & 3 & 8 & \text{yes}\\ 2 & 4 & 125/16 & \\ \hline 3 & 0 & 1 &\\ 3 & 1 & 8/3 & \\ 3 & 2 & 3 & \text{yes}\\ 3 & 3 & 64/27 & \\ \hline 5 & 0 & 1 & \\ 5 & 1 & 8/5 & \text{yes}\\ 5 & 2 & 27/25 & \\ \hline 7 & 0 & 1 & \\ 7 & 1 & 8/7 & \text{yes}\\ 7 & 2 & 27/49 & \\ \hline 11 & 0 & 1 & \text{yes} \\ 11 & 1 & 8/11 & \\ \hline \cdots & \cdots & \cdots & \end{array}\]

Finally, the number we seek is $N=2^3 3^2 5^1 7^1 = 2520.$ The sum of its digits is $2+5+2+0=\boxed{\textbf{(E) }9}.$

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

~MRENTHUSIASM

Solution 2 (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 $\textbf{(E) } 9$ works.

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

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

~ pi_is_3.14

See also

2021 AMC 12A (Problems • Answer Key • Resources)
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