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

Revision as of 20:53, 24 August 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 factorizations of $768$ and $384$ are $2^8\cdot3$ and $2^7\cdot3,$ respectively. Note that $f(n)$ is the sum of all fractions of the form $\frac 1d,$ where $d$ is a positive divisor of $n.$ By geometric series, it follows that $\begin{alignat*}{8} f(768)&=\left(\sum_{k=0}^{8}\frac{1}{2^k}\right)+\left(\sum_{k=0}^{8}\frac{1}{2^k\cdot3}\right)&&=\frac{511}{256}+\frac{511}{768}&&=\frac{2044}{768}, \\ f(384)&=\left(\sum_{k=0}^{7}\frac{1}{2^k}\right)+\left(\sum_{k=0}^{7}\frac{1}{2^k\cdot3}\right)&&=\frac{255}{128}+\frac{255}{384}&&=\frac{1020}{384}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

~lopkiloinm ~MRENTHUSIASM

Solution 2

The prime factorization of $384$ is $2^7\cdot3,$ so each of its positive divisors is of the form $2^m$ or $2^m\cdot3$ for some integer $m$ such that $0\leq m\leq7.$ We will use this fact to calculate the sum of all its positive divisors. Note that $2^m + 2^m\cdot3 = 2^m\cdot(1+3) = 2^m\cdot4 = 2^m\cdot2^2 = 2^{m+2}$ is the sum of the two forms of positive divisors for all such $m.$ By geometric series, the sum of all positive divisors of $384$ is $\sum_{k=2}^{9}2^k = \frac{2^{10}-2^2}{2-1} = 1020,$ from which $f(384) = \frac{1020}{384}.$ Similarly, since the prime factorization of $768$ is $2^8 \cdot 3,$ we have $f(768) = \frac{2044}{768}.$

Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

~mahaler

Solution 3

Let $\sigma(n)$ denotes the sum of all positive divisors of $n,$ so $f(n)=\sigma(n)\div n.$

Suppose that $n=\prod_{i=1}^{k}p_i^{e_i}$ is the prime factorization of $n.$ Since $\sigma(n)$ is multiplicative, we have $\sigma(n)=\sigma\left(\prod_{i=1}^{k}p_i^{e_i}\right)=\prod_{i=1}^{k}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{k}\left(\sum_{j=0}^{e_i}p_i^j\right)=\prod_{i=1}^{k}\frac{p_i^{e_i+1}-1}{p_i-1}$ by geometric series.

The prime factorizations of $768$ and $384$ are $2^8\cdot3$ and $2^7\cdot3,$ respectively. Note that $\begin{alignat*}{8} f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{2044}{768}, \\ f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{1020}{384}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

~MRENTHUSIASM

Solution 4 (Recursion)

$384 = 2^6\cdot3$

$768 = 2^7\cdot3$

Both numbers take the form $2^n\cdot3$ . Define a function $S(n)$ as being the sum of the factors for any number $2^n\cdot3$ (where $n$ is a nonnegative integer.)

If you list out the factors of a number $2^n\cdot3$ , all the factors are even except for $1$ and $3$ . So to produce a list of factors for the number $2^{n+1}\cdot3$ , you can multiply all the factors on the first list by $2$ and then add on $1$ and $3$ . Thus, $S(n) = 2S(n-1) + 4$ .

We need to find $\frac{S(7)}{768} - \frac{S(6)}{384}.$ This can be written as $\frac{S(7)-2S(6)}{768}$ . Our recursive formula states that $S(n) = 2S(n-1) + 4$ so $S(n) - 2S(n-1) = 4$ . Thus, the fraction simplifies to $\frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}.$

~Curious_crow

@@ Line 39: / Line 39: @@
 ~MRENTHUSIASM
+==Solution 4 (Recursion)==
+<math>384 = 2^6\cdot3</math>
+<math>768 = 2^7\cdot3</math>
+Both numbers take the form <math>2^n\cdot3</math>. Define a function <math>S(n)</math> as being the sum of the factors for any number <math>2^n\cdot3</math> (where <math>n</math> is a nonnegative integer.)
+If you list out the factors of a number <math>2^n\cdot3</math>, all the factors are even except for <math>1</math> and <math>3</math>. So to produce a list of factors for the number <math>2^{n+1}\cdot3</math>, you can multiply all the factors on the first list by <math>2</math> and then add on <math>1</math> and <math>3</math>. Thus, <math>S(n) = 2S(n-1) + 4</math>.
+We need to find <math>\frac{S(7)}{768} - \frac{S(6)}{384}.</math> This can be written as <math>\frac{S(7)-2S(6)}{768}</math>. Our recursive formula states that <math>S(n) = 2S(n-1) + 4</math> so <math>S(n) - 2S(n-1) = 4</math>. Thus, the fraction simplifies to <math>\frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}.</math>
+~Curious_crow
 ==See Also==
 {{AMC12 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 {{MAA Notice}}

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