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

Revision as of 18:06, 10 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 and and 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 $(0,0)$ , $(1,1)$ , $(-1,1)$ , and $(0,2)$ gives a total of $\boxed{\textbf{(D) }4}$ ordered pairs.

Solution 2

Bringing all of the terms to the LHS, we see a quadratic equation $y^2 - 2y + x^{2020} = 0$ in terms of $y$ . Applying the quadratic formula, we get $y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.$ In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, $4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.$ Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$ .

The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$ .

Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{\textbf{(D) }4}$ .

~Tiblis

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

@@ Line 6: / Line 6: @@
 ==Solution==
 Rearranging the terms and and 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>(1,1)</math>, <math>(-1,1)</math>, and <math>(0,2)</math> gives  a  total  of <math>\boxed{\textbf{(D) }4}</math> ordered pairs.
+==Solution 2==
+Bringing all of the terms to the LHS, we see a quadratic equation <cmath>y^2 - 2y + x^{2020} = 0</cmath> in terms of <math>y</math>. Applying the quadratic formula, we get <cmath>y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.</cmath> In order for <math>y</math> to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, <math>4(1-x^{2020})</math> must be nonnegative. Therefore, <cmath>4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.</cmath> Here, we see that we must split the inequality into a compound, resulting in <math>-1 \leq x \leq 1</math>.
+The only integers that satisfy this are <math>x \in \{-1,0,1\}</math>. Plugging these values back into the quadratic equation, we see that <math>x = \{-1,1\}</math> both produce a discriminant of <math>0</math>, meaning that there is only 1 solution for <math>y</math>. If <math>x = \{0\}</math>, then the discriminant is nonzero, therefore resulting in two solutions for <math>y</math>.
+Thus, the answer is <math>2 \cdot 1 + 1 \cdot 2 = \boxed{\textbf{(D) }4}</math>.
+~Tiblis
 ==Video Solution==

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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