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

(Created page with "==Problem== Let <math>x</math> and <math>y</math> be nonnegative integers such that <math>(x+y)^2+(xy)^2=25.</math> Find the sum of all possible values of <math>x.</math> ==S...")
 
(Solution)
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==Solution==
 
==Solution==
 
asdf
 
asdf
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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>
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13.
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Case 1: <math>x+y=4,3</math>.
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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>.
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Case 2: <math>x+y=5,0</math>
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This works when <math>x=0,5</math>.
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Therefore, the answer is <math>9</math>. ~ kante314

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

asdf

Easily, we can see that $C=3,5$. $C$ must be $5$ because $795$ is divisible by $3$ and $793$ is not divisible by $3$. Therefore $\frac{795}{3}=265$. So, $3A+2B+C=6+12+5=23$

13. 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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