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

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Case 2: <math>x+y=3, xy=4</math>.
 
Case 2: <math>x+y=3, xy=4</math>.
  
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 3 + 4</math>, which are not real.  
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 3x + 4</math>, which are not real.  
  
 
Case 3: <math>x+y=4,xy=3</math>.
 
Case 3: <math>x+y=4,xy=3</math>.
  
−
The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 4 + 3</math>, which are <math>1</math> and <math>3</math>.  
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 4x + 3</math>, which are <math>1</math> and <math>3</math>.  
  
 
Case 4: <math>x+y=5, xy = 0</math>
 
Case 4: <math>x+y=5, xy = 0</math>
  
−
The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 5</math>, which are <math>0</math> and <math>5</math>.  
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 5x</math>, which are <math>0</math> and <math>5</math>.  
  
 
Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>.  
 
Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>.  
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~Revised and Edited by Mathdreams
 
~Revised and Edited by Mathdreams
 +
 +
~Some Edits by BakedPotato66
  
 
==Solution 2==
 
==Solution 2==
−
Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4 \}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math>
+
Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4,5\}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math>
 
<math>\linebreak</math>
 
<math>\linebreak</math>
 
~Geometry285
 
~Geometry285
 +
 +
 +
 +
==See also==
 +
#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
 +
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 16:15, 12 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

Notice that since $x$ and $y$ are both integers, $x+y$ and $xy$ are also both integers. We can then use casework to determine all possible values of $x$:

Case 1: $x+y=0,xy=5$.

The solutions for $x$ and $y$ are the roots of $x^2 + 5$, which are not real.

Case 2: $x+y=3, xy=4$.

The solutions for $x$ and $y$ are the roots of $x^2 - 3x + 4$, which are not real.

Case 3: $x+y=4,xy=3$.

The solutions for $x$ and $y$ are the roots of $x^2 - 4x + 3$, which are $1$ and $3$.

Case 4: $x+y=5, xy = 0$

The solutions for $x$ and $y$ are the roots of $x^2 - 5x$, which are $0$ and $5$.

Therefore, the answer is $1 + 3 + 0 + 5 = 9$.


~kante314

~Revised and Edited by Mathdreams

~Some Edits by BakedPotato66

Solution 2

Note we are dealing with Pythagorean triples, so $xy=\{0,3,4,5\}$, and we have $x+y$ is a member of the set too. We see $x+y=4$ has $x=\{1,3 \}$ work, but $x+y=4$ has nothing work. If $x+y=0$, we have $x=\{0,5 \}$ work. The answer is $0+1+3+5=\boxed{9}$ $\linebreak$ ~Geometry285


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

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