Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME I Problems/Problem 11"

(→‎Solution)
(→‎Solution)
Line 8: Line 8:
  
 
Using the formula for a [[geometric series]], this reduces to <math>S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}</math>, and <math>\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}</math>.
 
Using the formula for a [[geometric series]], this reduces to <math>S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}</math>, and <math>\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}</math>.
−
 
−
 
−
*NOTE: Hi, I'm just someone wondering: isn't the question saying that S is the sum of fractions where a and b are relatively prime divisors of 1000? So for example 5/8 would be okay but 10/25 wouldn't. If that's the case, then a and b would only be 5, 25, 125, 2, 4, and 8. So S would be:
 
−
(5+25+125)/8 + (5+25+125)/4 + (5+25+125)/2 + (2+4+8)/5 + (2+4+8)/25 + (2+4+8)/125 = 434/125 + 1085/8
 
−
then our solution would be the largest integer less than (3+135)/10, which would be 13.
 
  
 
== See also ==
 
== See also ==

Revision as of 14:49, 21 January 2015

Problem

Let $S$ be the sum of all numbers of the form $a/b,$ where $a$ and $b$ are relatively prime positive divisors of $1000.$ What is the greatest integer that does not exceed $S/10$?

Solution

Since all divisors of $1000 = 2^35^3$ can be written in the form of $2^{m}5^{n}$, it follows that $\frac{a}{b}$ can also be expressed in the form of $2^{x}5^{y}$, where $-3 \le x,y \le 3$. Thus every number in the form of $a/b$ will be expressed one time in the product

\[(2^{-3} + 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^2 + 2^3)(5^{-3} + 5^{-2} +5^{-1} + 5^{0} + 5^{1} + 5^2 + 5^3)\]

Using the formula for a geometric series, this reduces to $S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}$, and $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$.

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_11&oldid=66909"