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

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<cmath>\frac{1 + 3}{768}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath>
 
<cmath>\frac{1 + 3}{768}=\boxed{\textbf{(B)}\ \frac{1}{192}}.</cmath>
 
~pengf
 
~pengf
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==Solution 6==
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As in Solution 1, the prime factorization of <math>768</math> is <math>2^8 \cdot 3</math>, and the prime factorization of <math>384</math> is <math>2^7 \cdot 3</math>. Recalling the sum of factors formula and sum of the powers of 2 formula, we find that the sum of the factors of <math>768</math> is <math>(1+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8)(1+3^1)=(2^9-1)(4)=(511)(4)</math>, and the sum of factors of <math>384</math> is <math>1+2^1+2^2+2^3+2^4+2^5+2^6+2^7)(1+3^1)=(2^8-1)(4)=(255)(4)</math>. Then <math>f(768)=\frac{(511)(4)}{768}=\frac{511}{192}</math>, and <math>f(384)=\frac{(255)(4)}{384}=\frac{510}{192}</math>. So, <math>f(768)-f(384)=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}</math>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 12:48, 4 April 2024

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

Note that \begin{align*} 384 &= 2^7\cdot3, \\ 768 &= 2^8\cdot3. \end{align*} 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 $\frac{S(8)}{768} - \frac{S(7)}{384}$. This can be written as $\frac{2S(7)+4-2S(7)}{768}$. Our recursive formula states that $S(n) = 2S(n-1) + 4$ so $S(n) - 2S(n-1) = 4$. Thus, the fraction simplifies to $\frac{4}{768} = \boxed{\textbf{(B)}\ \frac{1}{192}}$. Note that we never needed to figure out what $S(8)$ or $S(7)$ were, sparing us a lot of calculation.

~Curious_crow

Solution 5

We know that \begin{align*} f(768) &= \frac{1 + 2 + 3 + 4 + \cdots + 768}{768}, \\ f(384) &= \frac{1 + 2 + 3 + 4 + \cdots + 384}{384}. \end{align*} We want to find the difference of these so \[\frac{1 + 2 + 3 + 4 + \cdots + 768}{768} - \frac{ 2(1 + 2 + 3 + 4 + \cdots + 384)}{2\cdot384} = \frac{1 + 2 + 3 + 4 + \cdots + 768}{768} - \frac{ 2 + 4 + 6 + 8 + \cdots + 768}{768}.\] We are only left with the even divisors of $768$ on the fraction on the right so the answer would be the sum of all odd divisors of $768$ ($1$ and $3$) over $768$: \[\frac{1 + 3}{768}=\boxed{\textbf{(B)}\ \frac{1}{192}}.\] ~pengf

Solution 6

As in Solution 1, the prime factorization of $768$ is $2^8 \cdot 3$, and the prime factorization of $384$ is $2^7 \cdot 3$. Recalling the sum of factors formula and sum of the powers of 2 formula, we find that the sum of the factors of $768$ is $(1+2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8)(1+3^1)=(2^9-1)(4)=(511)(4)$, and the sum of factors of $384$ is $1+2^1+2^2+2^3+2^4+2^5+2^6+2^7)(1+3^1)=(2^8-1)(4)=(255)(4)$. Then $f(768)=\frac{(511)(4)}{768}=\frac{511}{192}$, and $f(384)=\frac{(255)(4)}{384}=\frac{510}{192}$. So, $f(768)-f(384)=\frac{511}{192}-\frac{510}{192}=\boxed{\textbf{(B)}\ \frac{1}{192}}$.

~ cxsmi

Video Solution by OmegaLearn

https://youtu.be/jgyyGeEKhwk?t=1116

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/FFlj8W_xkPc

~IceMatrix

See Also

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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