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2021 Fall AMC 12B Problems/Problem 12

Revision as of 22:26, 12 January 2022 by Mahaler (talk | contribs) (→‎Solution 1)

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 $2^8*3$ and the prime factorization of $384$ is $2^7*3$ so \[f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}\] \[f(384)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}\] so the difference is $\boxed{\textbf{(B)}\ \frac{1}{192}}$ ~lopkiloinm

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

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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