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

Revision as of 00:21, 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 $\begin{alignat*}{8} f(768)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\ f(384)&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\cdots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\cdots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{85}{32}. \end{alignat*}$ 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

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 $7.$ Therefore, 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 $f(768) = \frac{2044}{768}.$

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

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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