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

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=== Solution 6: Simplest==
 
=== Solution 6: Simplest==
 
We see that <math>x</math> has to be <math>1</math>, <math>0</math>, or <math>-1</math>, as any other integer would make this value too large. We also know that because <math>2020</math> is even, both <math>-1</math>, and <math>1</math> for <math>x</math> will yield the same <math>x^{2020}</math> value of <math>1</math>.  
 
We see that <math>x</math> has to be <math>1</math>, <math>0</math>, or <math>-1</math>, as any other integer would make this value too large. We also know that because <math>2020</math> is even, both <math>-1</math>, and <math>1</math> for <math>x</math> will yield the same <math>x^{2020}</math> value of <math>1</math>.  
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'''Case 1''': <math>x=0</math>. This gives us that <math>y^2 = 2y</math>. Dividing both sides by <math>y</math> gives us <math>y = 2</math>. Additionally, we know intuitively that <math>y = 0</math> is also a case, which gives us 2 possibilities for this case.
 
'''Case 1''': <math>x=0</math>. This gives us that <math>y^2 = 2y</math>. Dividing both sides by <math>y</math> gives us <math>y = 2</math>. Additionally, we know intuitively that <math>y = 0</math> is also a case, which gives us 2 possibilities for this case.
 
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<ol style="margin-left: 1.5em;">

Revision as of 19:43, 4 November 2022

The following problem is from both the 2020 AMC 10B #9 and 2020 AMC 12B #8, so both problems redirect to this page.

Problem

How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$

Solutions

Solution 1

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: Solve for 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 the discriminant must be 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 $x=0$: \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$.

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

~edits by BakedPotato66

Solution 4: Solve for 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 $(0,2)$, $(0,0)$, $(1,1)$, $(-1,1)$. These are the only $4$ solutions, so the answer is $\boxed{\textbf{(D) } 4}$

- wwt7535

~ edits by BakedPotato66

Solution 5: Similar to solution 4

Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$, which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$.

For $y=0$, $x$ can only be $0$.

For $y=1$, $x^2=1$ so $x=1, -1$.

For $y=2$, $x$ can only be $0$ as well.

This gives us the solutions $(0, 0)$, $(1, 1)$, $(-1, 1)$, and $(0, 2)$. These are the only solutions, so there is a total of $\boxed{\textbf{(D) } 4}$ ordered pairs.

- kc1374

= Solution 6: Simplest

We see that $x$ has to be $1$, $0$, or $-1$, as any other integer would make this value too large. We also know that because $2020$ is even, both $-1$, and $1$ for $x$ will yield the same $x^{2020}$ value of $1$.

    Case 1: $x=0$. This gives us that $y^2 = 2y$. Dividing both sides by $y$ gives us $y = 2$. Additionally, we know intuitively that $y = 0$ is also a case, which gives us 2 possibilities for this case.
      Case 2: $x = 1$ or $-1$. This gives us that $1+y^ =2 y$. Bringing the $2y$ to the other side, we have a simple quadratic. $y^2-2y + 1 = 0$ = $(y-1)^2$ so $y = 1$. Because this works for $x$ as $-1$ and $1$, there are 2 possibilities for this case. Adding the cases gets us our final answer of $\boxed{\textbf{(D) } 4}$ ordered pairs. ~iluvme

      Video Solutions

      Video Solution 1

      https://youtu.be/6ujfjGLzVoE

      ~IceMatrix

      Video Solution 2

      https://youtu.be/7dQ423hhgac

      ~savannahsolver

      Video Solution 3

      https://youtu.be/zfChnbMGLVQ?t=4251

      ~pi_is_3.14

      Video Solution 4

      https://youtu.be/FfATkdxncG4

      ~Education, the Study of Everything

      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
      2020 AMC 12B (Problems • Answer Key • Resources)
      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

      The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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