Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"

Revision as of 19:09, 26 January 2022

Problem

For $n$ a positive integer, let $f(n)$ be the quotient obtained when the sum of all positive divisors of n is divided by n. For example, $f(14)=(1+2+7+14)\div 14=\frac{12}{7}$ What is $f(768)-f(384)?$

$\textbf{(A)}\ \frac{1}{768} \qquad\textbf{(B)}\ \frac{1}{192} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\ \frac{4}{3} \qquad\textbf{(E)}\ \frac{8}{3}$

Solution 1

The prime factorization of $768$ is $2^8\cdot3,$ and the prime factorization of $384$ is $2^7\cdot3.$ Note that $\begin{alignat*}{8} f(768)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\ f(384)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{85}{32}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.$ ~lopkiloinm ~MRENTHUSIASM

Solution 2

Let $\sigma(n)$ denotes the sum of the positive integer divisors of $n,$ so $f(n)=\frac{\sigma(n)}{n}.$

Suppose that $n=\prod_{i=1}^{k}p_i^{e_i}$ is the prime factorization of $n.$ Since $\sigma(n)$ is multiplicative, we have $\sigma(n)=\sigma\left(\prod_{i=1}^{k}p_i^{e_i}\right)=\prod_{i=1}^{k}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{k}\left(\sum_{j=0}^{e_i}p_i^j\right)=\prod_{i=1}^{k}\frac{p_i^{e_i+1}-1}{p_i-1}.$ The prime factorization of $768$ is $2^8\cdot3,$ and the prime factorization of $384$ is $2^7\cdot3.$ Note that $\begin{alignat*}{8} f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{511}{192}, \\ f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{85}{32}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.$ ~MRENTHUSIASM

Solution 3

We see that the prime factorization of $384$ is $2^7 \cdot 3.$ Each of its divisors is in the form of $2^x$ or $2^x \cdot 3$ for a nonnegative integer $x \le 7.$ We can use this fact to our advantage when calculating the sum of all of them. Notice that $2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}$ is the sum of the two forms of divisors for each $x$ from $0$ to $7,$ inclusive. So, the sum of all of the divisors of $384$ is just $2^2 + 2^3 + 2^4 + \cdots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \cdots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020.$ Therefore, we have $f(384) = \frac{1020}{384}.$ Similarly, since $768 = 2^8 \cdot 3,$ we have $f(768) = \frac{2044}{768}.$

Finally, the answer is $f(768) - f(384) = \frac{2044}{768} - \frac{1020}{384} = \frac{2044}{768} - \frac{2040}{768} = \frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}.$

~mahaler

@@ Line 12: / Line 12: @@
 The prime factorization of <math>768</math> is <math>2^8\cdot3,</math> and the prime factorization of <math>384</math> is <math>2^7\cdot3.</math> Note that
 <cmath>\begin{alignat*}{8}
-f(768)&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\ldots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\
+f(768)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\
-f(384)&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\ldots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{510}{192}.
+f(384)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{85}{32}.
 \end{alignat*}</cmath>
-Therefore, the answer is <math>f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.</math>
+Therefore, the answer is <cmath>f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath>
 ~lopkiloinm ~MRENTHUSIASM
 ==Solution 2==
-We see that the prime factorization of <math>384</math> is <math>2^7 \cdot 3</math>. Each of its divisors is in the form of <math>2^x</math> or <math>2^x \cdot 3</math> for a nonnegative integer <math>x \le 7</math>. We can use this fact to our advantage when calculating the sum of all of them. Notice that <math>2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}</math> is the sum of the two forms of divisors for each <math>x</math> from <math>0-7</math>, inclusive. So, the sum of all of the divisors of <math>384</math> is just <math>2^2 + 2^3 + 2^4 + \ldots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \ldots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020</math>. Therefore, <math>f(384) = \frac{1020}{384}</math>. Similarly, since <math>768 = 2^8 \cdot 3</math>, <math>f(768) = \frac{2044}{768}</math>. Therefore, the answer is <math>f(768) - f(384) = \frac{2044}{768} - \frac{1020}{384} = \frac{2044}{768} - \frac{2040}{768} = \frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}</math>.
+Let <math>\sigma(n)</math> denotes the sum of the positive integer divisors of <math>n,</math> so <math>f(n)=\frac{\sigma(n)}{n}.</math>
+Suppose that <math>n=\prod_{i=1}^{k}p_i^{e_i}</math> is the prime factorization of <math>n.</math> Since <math>\sigma(n)</math> is multiplicative, we have <cmath>\sigma(n)=\sigma\left(\prod_{i=1}^{k}p_i^{e_i}\right)=\prod_{i=1}^{k}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{k}\left(\sum_{j=0}^{e_i}p_i^j\right)=\prod_{i=1}^{k}\frac{p_i^{e_i+1}-1}{p_i-1}.</cmath>
+The prime factorization of <math>768</math> is <math>2^8\cdot3,</math> and the prime factorization of <math>384</math> is <math>2^7\cdot3.</math> Note that
+<cmath>\begin{alignat*}{8}
+f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{511}{192}, \\
+f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{85}{32}.
+\end{alignat*}</cmath>
+Therefore, the answer is <cmath>f(768)-f(384)=\frac{511}{192}-\frac{85}{32}=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath>
+~MRENTHUSIASM
+==Solution 3==
+We see that the prime factorization of <math>384</math> is <math>2^7 \cdot 3.</math> Each of its divisors is in the form of <math>2^x</math> or <math>2^x \cdot 3</math> for a nonnegative integer <math>x \le 7.</math> We can use this fact to our advantage when calculating the sum of all of them. Notice that <cmath>2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}</cmath> is the sum of the two forms of divisors for each <math>x</math> from <math>0</math> to <math>7,</math> inclusive. So, the sum of all of the divisors of <math>384</math> is just <cmath>2^2 + 2^3 + 2^4 + \cdots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \cdots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020.</cmath> Therefore, we have <math>f(384) = \frac{1020}{384}.</math> Similarly, since <math>768 = 2^8 \cdot 3,</math> we have <math>f(768) = \frac{2044}{768}.</math>
+Finally, the answer is <cmath>f(768) - f(384) = \frac{2044}{768} - \frac{1020}{384} = \frac{2044}{768} - \frac{2040}{768} = \frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath>
 ~mahaler

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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