2021 Fall AMC 12B Problems/Problem 12

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 $\begin{alignat*}{8} f(768)&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\ldots+\frac{1}{2^8\cdot3}\right)&&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)\left(1+\frac{1}{3}\right)&&=\frac{511}{256}\cdot\frac{4}{3}&&=\frac{511}{192}, \\ f(384)&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^8}\right)+\left(\frac{1}{3}+\frac{1}{2\cdot3}+\ldots+\frac{1}{2^7\cdot3}\right)&&=\left(1+\frac{1}{2}+\ldots+\frac{1}{2^7}\right)\left(1+\frac{1}{3}\right)&&=\frac{255}{128}\cdot\frac{4}{3}&&=\frac{510}{192}. \end{alignat*}$ Therefore, the answer is $f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.$

~lopkiloinm ~MRENTHUSIASM

Solution 2

We see that the prime factorization of $384$ is $2^7 \cdot 3$ . Each of its divisors is in the form of $2^x$ or $2^x \cdot 3$ for a nonnegative integer $x \le 7$ . We can use this fact to our advantage when calculating the sum of all of them. Notice that $2^x + 2^x \cdot 3 = 2^x(1+3) = 2^x \cdot 4 = 2^x \cdot 2^2 = 2^{x+2}$ is the sum of the two forms of divisors for each $x$ from $0-7$ , inclusive. So, the sum of all of the divisors of $384$ is just $2^2 + 2^3 + 2^4 + \ldots + 2^9 = (2^0 + 2^1 + 2^2 + 2^3 + 2^4 + \ldots + 2^9) - (2^0 + 2^1) = (2^{10} - 1) - (2^0 + 2^1) = 1020$ . Therefore, $f(384) = \frac{1020}{384}$ . Similarly, since $768 = 2^8 \cdot 3$ , $f(768) = \frac{2044}{768}$ . Therefore, 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

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

Contents

Problem

Solution 1

Solution 2

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