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

AoPS Wiki talk:Problem of the Day/July 21, 2011

Revision as of 16:02, 21 July 2011 by Brianyang (talk | contribs) (→‎Solution)

Problem

AoPSWiki:Problem of the Day/July 21, 2011

Solution

Note: someone check the arithmetic please

For purposes of generalization, let the equations be $ax+by=1002001$ and $cx+dy=2004002$. Notice that $2*1002001 = 2004002$. Swap this out for a new variable $z$. This gives $ax+by=z, cx+dy=2z$.

Multiply the left equation by two and substitute it into the other equation.

$2ax+2by=cx+dy$, which implies that$(c-2a)x=(2b-d)y$. Substituting the actual numbers back in gives: $(997997-686686)x=(630630-8008)y$ Simplifying: $311311x=622722y$ Which further simplifies to: $x=2y$

Then we just plug this back into the original parameter, where $z=1002001$.

$a(2y)+by=1002001$ $2ay+by=1002001$ $y(2a+b)=1002001$ $y=\frac{1002001}{2a+b}$ $=\frac{143}{53}$

Then this also means that $x=\frac{286}{106}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/July_21,_2011&oldid=40583"