Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"

(Solution)
(Solution)
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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 00:19, 11 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

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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

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