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 "2017 AIME I Problems/Problem 2"

(wrote the solution)
 
m
Line 14: Line 14:
  
 
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{062}</math>
 
The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{062}</math>
 +
 +
~IYN~

Revision as of 15:53, 8 March 2017

Problem 2

When each of $702$, $787$, and $855$ is divided by the positive integer $m$, the remainder is always the positive integer $r$. When each of $412$, $722$, and $815$ is divided by the positive integer $n$, the remainder is always the positive integer $s \neq r$. Find $m+n+r+s$.

Solution

Let's tackle the first part of the problem first. We can safely assume: \[702 = xm + r\] \[787 = ym + r\] \[855 = zm + r\] Now, if we subtract two values: \[787-702 = 85 = 17\cdot5\] which also equals \[(ym+r)-(xm+r) = m\cdot(y-x)\] Similarly, \[855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)\] Since $17$ is the only common factor, we can assume that $m=17$, and through simple division, that $r=5$.

Using the same method on the second half: \[412 = an + s\] \[722 = bn + s\] \[815 = cn + s\] Then. \[722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)\] \[815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)\] The common factor is $31$, so $n=31$ and through division, $s=9$.

The answer is $m+n+r+s = 17+31+5+9 = \boxed{062}$

~IYN~

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_2&oldid=84465"