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 "2021 JMPSC Accuracy Problems/Problem 13"

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
 
asdf
 
asdf
  
−
Easily, we can see that <math>C=3,5</math>. <math>C</math> must be <math>5</math> because <math>795</math> is divisible by <math>3</math> and <math>793</math> is not divisible by <math>3</math>. Therefore <math>\frac{795}{3}=265</math>. So, <math>3A+2B+C=6+12+5=23</math>
 
−
 
−
13.
 
 
Case 1: <math>x+y=4,3</math>.
 
Case 1: <math>x+y=4,3</math>.
 
There are no possible answer when <math>x+y=3</math>, but when <math>x+y=4</math>, <math>x</math> can equal <math>3</math> or <math>1</math>.
 
There are no possible answer when <math>x+y=3</math>, but when <math>x+y=4</math>, <math>x</math> can equal <math>3</math> or <math>1</math>.

Revision as of 23:19, 10 July 2021

Problem

Let $x$ and $y$ be nonnegative integers such that $(x+y)^2+(xy)^2=25.$ Find the sum of all possible values of $x.$

Solution

asdf

Case 1: $x+y=4,3$. There are no possible answer when $x+y=3$, but when $x+y=4$, $x$ can equal $3$ or $1$. Case 2: $x+y=5,0$ This works when $x=0,5$. Therefore, the answer is $9$. ~ kante314

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_13&oldid=157813"