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

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<math>\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9</math>
 
<math>\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9</math>
 
== Video Solution ==
 
https://www.youtube.com/watch?v=gWaUNz0gLE0
 
  
 
==Solution 1==
 
==Solution 1==
Consider the prime factorization <cmath>n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.</cmath> By the Multiplication Principle, <cmath>d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>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).</cmath> As <math>f(n)>0</math> for all positive integers <math>n,</math> it follows that for all positive integers <math>a</math> and <math>b</math>, <math>f(a)>f(b)</math> if and only if <math>f(a)^3>f(b)^3.</math> So, <math>f(n)</math> is maximized if and only if <cmath>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)</cmath> is maximized.
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Consider the prime factorization <cmath>n=\prod_{i=1}^{k}p_i^{e_i}.</cmath> By the Multiplication Principle, <cmath>d(n)=\prod_{i=1}^{k}(e_i+1).</cmath> Now, we rewrite <math>f(n)</math> as <cmath>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}}.</cmath> As <math>f(n)>0</math> for all positive integers <math>n,</math> it follows that for all positive integers <math>a</math> and <math>b, f(a)>f(b)</math> if and only if <math>f(a)^3>f(b)^3.</math> So, <math>f(n)</math> is maximized if and only if <cmath>f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}</cmath> is maximized.
  
For every factor <math>\frac{(e_i+1)^3}{{p_i}^{e_i}}</math> with a fixed <math>p_i</math> where <math>1\leq i\leq k,</math> the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime <math>p_i=2,3,5,7,\cdots,</math> we look for the <math>e_i</math> for which <math>\frac{(e_i+1)^3}{{p_i}^{e_i}}</math> is a relative maximum:
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For every independent factor <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> with a fixed <math>p_i,</math> where <math>1\leq i\leq k,</math> the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime <math>p_i</math> with <math>\left(p_1,p_2,p_3,p_4,\cdots\right)=\left(2,3,5,7,\cdots\right),</math> we look for the <math>e_i</math> for which <math>\frac{(e_i+1)^3}{p_i^{e_i}}</math> is a relative maximum:
<cmath>\begin{array}{ c c c c }
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<cmath>\begin{array}{c|c|c|c|c}  
p_i & e_i & (e_i+1)^3/\left({p_i}^{e_i}\right) & \text{max?} \\  
+
& & & & \\ [-2.25ex]
\hline\hline
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\boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{(e_i+1)^3/\left(p_i^{e_i}\right)} & \textbf{Max?} \\ [0.5ex]
2 & 0 & 1 & \\
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\hline\hline  
2 & 1 & 4 & \\
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& & & & \\ [-2ex]
2 & 2 & 27/4 &\\
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1 & 2 & 0 & 1 & \\    
2 & 3 & 8 & \text{yes}\\
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& & 1 & 4 & \\  
2 & 4 & 125/16 & \\
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& & 2 & 27/4 &\\  
\hline
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& & 3 & 8 & \checkmark\\  
3 & 0 & 1 &\\
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& & 4 & 125/16 & \\ [0.5ex]
3 & 1 & 8/3 & \\
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\hline
3 & 2 & 3 &  \text{yes}\\
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& & & & \\ [-2ex]
3 & 3 & 64/27 &  \\
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2 & 3 & 0 & 1 &\\  
\hline
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& & 1 & 8/3 & \\  
5 & 0 & 1 &  \\
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& & 2 & 3 &  \checkmark\\  
5 & 1 & 8/5 &  \text{yes}\\
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& & 3 & 64/27 &  \\ [0.5ex]
5 & 2 & 27/25 & \\
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\hline
\hline
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& & & & \\ [-2ex]
7 & 0 & 1 &  \\
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3 & 5 & 0 & 1 &  \\  
7 & 1 & 8/7 &  \text{yes}\\
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& & 1 & 8/5 &  \checkmark\\  
7 & 2 & 27/49 & \\
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& & 2 & 27/25 & \\ [0.5ex]
\hline
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\hline
11 & 0 & 1 & \text{yes} \\
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& & & & \\ [-2ex]
11 & 1 & 8/11 &  \\
+
4 & 7 & 0 & 1 &  \\  
\hline
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& & 1 & 8/7 &  \checkmark\\  
\cdots & \cdots & \cdots &
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& & 2 & 27/49 & \\ [0.5ex]
 +
\hline
 +
& & & & \\ [-2ex]
 +
\geq5 & \geq11 & 0 & 1 & \checkmark \\  
 +
& & \geq1 & \leq8/11 &   \\ [0.5ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
  
Finally, the number we seek is <math>N=2^3 3^2 5^1 7^1 = 2520.</math> The sum of its digits is <math>2+5+2+0=\boxed{\textbf{(E) }9}.</math>
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Finally, the positive integer we seek is <math>N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.</math> The sum of its digits is <math>2+5+2+0=\boxed{\textbf{(E) }9}.</math>
  
Actually, once we get that <math>3^2</math> is a factor of <math>N,</math> we know that the sum of the digits of <math>N</math> must be a multiple of <math>9.</math> Only choice <math>\textbf{(E)}</math> is possible.
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Actually, once we notice that <math>3^2</math> is a factor of <math>N,</math> we can conclude that the sum of the digits of <math>N</math> must be a multiple of <math>9.</math> Only choice <math>\textbf{(E)}</math> is possible.
 
    
 
    
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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-scrabbler94
 
-scrabbler94
  
== Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==
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== Video Solutions ==
https://youtu.be/6P-0ZHAaC_A
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https://www.youtube.com/watch?v=gWaUNz0gLE0 (by Dedekind Cuts)
 +
 
 +
https://www.youtube.com/watch?v=Sv4gj1vMjOs (by Aaron He)
 +
 
 +
https://youtube.com/watch?v=y_7s8fvMCdI (by Punxsutawney Phil)
 +
 
 +
https://youtu.be/6P-0ZHAaC_A (by OmegaLearn)
 +
 
  
 
~ pi_is_3.14
 
~ pi_is_3.14

Revision as of 22:21, 22 March 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 \[n=\prod_{i=1}^{k}p_i^{e_i}.\] By the Multiplication Principle, \[d(n)=\prod_{i=1}^{k}(e_i+1).\] 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 $n,$ 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 \[f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}\] is maximized.

For every independent 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 $p_i$ with $\left(p_1,p_2,p_3,p_4,\cdots\right)=\left(2,3,5,7,\cdots\right),$ we look for the $e_i$ for which $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: \[\begin{array}{c|c|c|c|c}  & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{(e_i+1)^3/\left(p_i^{e_i}\right)} & \textbf{Max?} \\ [0.5ex] \hline\hline  & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\      & & 1 & 4 & \\     & & 2 & 27/4 &\\     & & 3 & 8 & \checkmark\\     & & 4 & 125/16 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\     & & 1 & 8/3 & \\     & & 2 & 3 &  \checkmark\\     & & 3 & 64/27 &  \\ [0.5ex] \hline   & & & & \\ [-2ex] 3 & 5 & 0 & 1 &  \\     & & 1 & 8/5 &  \checkmark\\     & & 2 & 27/25 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 4 & 7 & 0 & 1 &  \\     & & 1 & 8/7 &  \checkmark\\     & & 2 & 27/49 & \\ [0.5ex] \hline   & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\     & & \geq1 & \leq8/11 &   \\ [0.5ex] \end{array}\]

Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{\textbf{(E) }9}.$

Actually, 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 (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

Video Solutions

https://www.youtube.com/watch?v=gWaUNz0gLE0 (by Dedekind Cuts)

https://www.youtube.com/watch?v=Sv4gj1vMjOs (by Aaron He)

https://youtube.com/watch?v=y_7s8fvMCdI (by Punxsutawney Phil)

https://youtu.be/6P-0ZHAaC_A (by OmegaLearn)


~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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Problem 24
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