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

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~Tiblis
 
~Tiblis
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==Solution 3, x first==
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Set it up as a quadratic in terms of y:
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<cmath>y^2-2y+x^{2020}=0</cmath>
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Then the discriminant is
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<cmath>\Delta = 4-4x^{2020}</cmath>
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This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive.
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Then <math>x=-1,0,1</math>. Checking each one:
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<math>-1</math> and <math>1</math> are the same when raised to the 2020th power:
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<cmath>y^2-2y+1=(y-1)^2=0</cmath>
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This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.
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Next, if <math>x=0</math>:
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<cmath>y^2-2y=0</cmath>
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Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math>
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These are the only 4 solutions, so <math>\boxed{\textbf{(D) } 4}</math>
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==Solution 4, y first==
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Move the <math>y^2</math> term to the other side to get <math>x^{2020}=2y-y^2 = y(2-y)</math>. Because <math>x^{2020} \geq 0</math> for all <math>x</math>, then <math>y(2-y) \geq 0 \Rightarrow y = 0,1,2</math>. If <math>y=0</math> or <math>y=2</math>, the right side is <math>0</math> and therefore <math>x=0</math>. When <math>y=1</math>, the right side become <math>1</math>, therefore <math>x=1,-1</math>. Our solutions are <math>(0,2)</math>, <math>(0,0)</math>, <math>(1,1)</math>, <math>(-1,1)</math>. There are <math>4</math> solutions, so the answer is <math>\boxed{\textbf{(D) } 4}</math>
  
 
==Video Solution==
 
==Video Solution==
https://youtu.be/OHR_6U686Qg
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https://youtu.be/6ujfjGLzVoE
  
 
~IceMatrix
 
~IceMatrix
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{{AMC10 box|year=2020|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2020|ab=B|num-b=8|num-a=10}}
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{{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}
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{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:28, 12 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


Solution 3, x first

Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $x^{2020} \leq 1$, because it is always positive. Then $x=-1,0,1$. Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$, so $(\pm 1,1)$ are solutions. Next, if $x=0$: \[y^2-2y=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$

These are the only 4 solutions, so $\boxed{\textbf{(D) } 4}$

Solution 4, y first

Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$. Because $x^{2020} \geq 0$ for all $x$, then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$. If $y=0$ or $y=2$, the right side is $0$ and therefore $x=0$. When $y=1$, the right side become $1$, therefore $x=1,-1$. Our solutions are $(0,2)$, $(0,0)$, $(1,1)$, $(-1,1)$. There are $4$ solutions, so the answer is $\boxed{\textbf{(D) } 4}$

Video Solution

https://youtu.be/6ujfjGLzVoE

~IceMatrix

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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
2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 12 Problems and Solutions

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