Maths Olympiad Prep

Track / Stage 3 / 216 of 260 #216 of 1964

Problem 216

AMC 10/12, early questions
Number theory Difficulty 3.8 Multiple choice

For nn a positive integer, let f(n)f(n) be the quotient obtained when the sum of all positive divisors of nn is divided by n.n. For example, f(14)=(1+2+7+14)÷14=127f(14)=(1+2+7+14)\div 14=\frac{12}{7}
What is f(768)f(384)?f(768)-f(384)?

Pick one

Official solution

The prime factorizations of 768768 and 384384 are 2832^8\cdot3 and 273,2^7\cdot3, respectively. Note that f(n)f(n) is the sum of all fractions of the form 1d,\frac 1d, where dd is a positive divisor of n.n. By geometric series, it follows that
f(768)=(k=0812k)+(k=0812k3)=511256+511768=2044768,f(384)=(k=0712k)+(k=0712k3)=255128+255384=1020384.\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)=(B) 1192.f(768)-f(384)=\boxed{\textbf{(B)}\ \frac{1}{192}}.
~lopkiloinm ~MRENTHUSIASM

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.