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

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==Solution 4 (Recursion)==
 
==Solution 4 (Recursion)==
  
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<math>384 = 2^6\cdot3</math>
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<math>384 = 2^7\cdot3</math>
  
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<math>768 = 2^7\cdot3</math>
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<math>768 = 2^8\cdot3</math>
  
  
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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> Note that we never needed to figure out what <math>S(7)</math> or <math>S(6)</math> were, sparing us a lot of calculation.
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We need to find <math>\frac{S(8)}{768} - \frac{S(7)}{384}.</math> This can be written as <math>\frac{S(8)-2S(8)}{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> Note that we never needed to figure out what <math>S(8)</math> or <math>S(7)</math> were, sparing us a lot of calculation.
  
 
~Curious_crow
 
~Curious_crow

Revision as of 22:21, 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)

$384 = 2^7\cdot3$

$768 = 2^8\cdot3$


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 $\frac{S(8)}{768} - \frac{S(7)}{384}.$ This can be written as $\frac{S(8)-2S(8)}{768}$. 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

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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