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

Revision as of 12:46, 27 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 $\begin{alignat*}{8} f(768)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{2044}{768}, \\ f(384)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{1020}{384}. \end{alignat*}$ 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 in the form $2^m$ or $2^m\cdot3$ for some nonnegative integer $m\leq7.$ 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 each $m$ from $0$ through $7.$ By geometric series, the sum of all positive divisors of $384$ is $2^2 + 2^3 + 2^4 + \cdots + 2^9 = \frac{2^{10}-2^2}{2-1} = 1020,$ 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 the positive integer 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 $\sigma(n)=\sigma\left(\prod_{i=1}^{k}p_i^{e_i}\right)=\prod_{i=1}^{k}\sigma\left(p_i^{e_i}\right)=\prod_{i=1}^{k}\left(\sum_{j=0}^{e_i}p_i^j\right)=\prod_{i=1}^{k}\frac{p_i^{e_i+1}-1}{p_i-1}.$ The prime factorization of $768$ is $2^8\cdot3,$ and the prime factorization of $384$ is $2^7\cdot3.$ Note that $\begin{alignat*}{8} f(768) &= \left(\frac{2^9-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div768 &&= \frac{2044}{768}, \\ f(384) &= \left(\frac{2^8-1}{2-1}\cdot\frac{3^2-1}{3-1}\right)\div384 &&= \frac{1020}{384}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

~MRENTHUSIASM

@@ Line 27: / Line 27: @@
 ==Solution 3==
-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>
+Let <math>\sigma(n)</math> denotes the sum of the positive integer divisors of <math>n,</math> so <math>f(n)=\sigma(n)\div 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>

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

Page

Toolbox

Search