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Difference between revisions of "2020 AMC 10B Problems/Problem 9"

(Created page with "==Problem== How many ordered pairs of integers <math>(x, y)</math> satisfy the equation <cmath>x^{2020}+y^2=2y?</cmath> <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\tex...")
 
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==Solution==
 
==Solution==
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Rearranging the terms in the equation gives <math>x^{2020}+y^2-2y=0</math>. Completing the square for <math>y</math> yields the result <math>x^{2020}+(y-1)^2=1</math>. Then, notice that <math>x</math> can only be <math>0</math>, <math>1</math> and <math>-1</math> because any value of <math>x^2020</math> that is greater than 1 will cause the term <math>(y-1)^2</math> to be less than <math>0</math>, which is impossible as <math>y</math> must be real. Therefore, plugging in the above values for <math>x</math> gives the ordered pairs <math>(0,0)</math>, <math>(0,2)</math>, <math>(1,-1)</math>, and <math>(-1,-1)</math>.
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Rearranging the terms in the equation gives <math>x^{2020}+y^2-2y=0</math>. Completing the square for <math>y</math> yields the result <math>x^{2020}+(y-1)^2=1</math>. Then, notice that <math>x</math> can only be <math>0</math>, <math>1</math> and <math>-1</math> because any value of <math>x^{2020}</math> that is greater than 1 will cause the term <math>(y-1)^2</math> to be less than <math>0</math>, which is impossible as <math>y</math> must be real. Therefore, plugging in the above values for <math>x</math> gives the ordered pairs <math>(0,0)</math>, <math>(0,2)</math>, <math>(1,-1)</math>, and <math>(-1,-1)</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:43, 7 February 2020

Problem

How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\] $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solution

Rearranging the terms in the equation gives $x^{2020}+y^2-2y=0$. Completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$. Then, notice that $x$ can only be $0$, $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$, which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$, $(0,2)$, $(1,-1)$, and $(-1,-1)$.

Video Solution

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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