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

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==Problem==
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{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #9]] and [[2020 AMC 12B Problems|2020 AMC 12B #8]]}}
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== Problem ==
  
 
How many ordered pairs of integers <math>(x, y)</math> satisfy the equation <cmath>x^{2020}+y^2=2y?</cmath>
 
How many ordered pairs of integers <math>(x, y)</math> satisfy the equation <cmath>x^{2020}+y^2=2y?</cmath>
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<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}</math>
 
<math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}</math>
  
==Solution==
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== Solutions ==
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=== Solution 1 ===
 
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.
 
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.
  
 
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=== Solution 2 ===
==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>.  
 
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>.  
  
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~Tiblis
 
~Tiblis
  
 
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=== Solution 3: Solve for x first ===
==Solution 3, x first==
 
 
Set it up as a quadratic in terms of y:
 
Set it up as a quadratic in terms of y:
 
<cmath>y^2-2y+x^{2020}=0</cmath>
 
<cmath>y^2-2y+x^{2020}=0</cmath>
 
Then the discriminant is
 
Then the discriminant is
 
<cmath>\Delta = 4-4x^{2020}</cmath>
 
<cmath>\Delta = 4-4x^{2020}</cmath>
This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive.
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This will clearly only yield real solutions when <math>|x^{2020}| \leq 1</math>, because the discriminant must be positive.
 
Then <math>x=-1,0,1</math>. Checking each one:
 
Then <math>x=-1,0,1</math>. Checking each one:
 
<math>-1</math> and <math>1</math> are the same when raised to the 2020th power:
 
<math>-1</math> and <math>1</math> are the same when raised to the 2020th power:
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This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.
 
This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions.
 
Next, if <math>x=0</math>:
 
Next, if <math>x=0</math>:
<cmath>y^2-2y=0</cmath>
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<cmath>y^2-2y=0 \Rightarrow y(y-2)=0</cmath>
Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math>
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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 our answer is <math>\boxed{\textbf{(D) } 4}</math>.
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~edits by BakedPotato66
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=== Solution 4: Solve for 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>.
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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>.
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If <math>y=0</math> or <math>y=2</math>, the right side is <math>0</math> and therefore <math>x=0</math>.
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When <math>y=1</math>, the right side become <math>1</math>, therefore <math>x=1,-1</math>.
  
These are the only 4 solutions, so <math>\boxed{\textbf{(D) } 4}</math>
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Our solutions are <math>(0,2)</math>, <math>(0,0)</math>, <math>(1,1)</math>, <math>(-1,1)</math>. These are the only <math>4</math> solutions, so the answer is <math>\boxed{\textbf{(D) } 4}</math>  
  
==Solution 4, y first==
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- wwt7535
  
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>
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~ edits by BakedPotato66
  
==Video Solution==
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== Video Solutions ==
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===Video Solution 1===
 
https://youtu.be/6ujfjGLzVoE
 
https://youtu.be/6ujfjGLzVoE
  
 
~IceMatrix
 
~IceMatrix
  
==See Also==
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===Video Solution 2===
 +
https://youtu.be/7dQ423hhgac
 +
 
 +
~savannahsolver
 +
 
 +
===Video Solution 3===
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https://youtu.be/zfChnbMGLVQ?t=4251
 +
 
 +
~ pi_is_3.14
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== See Also ==
  
 
{{AMC10 box|year=2020|ab=B|num-b=8|num-a=10}}
 
{{AMC10 box|year=2020|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2020|ab=B|num-b=7|num-a=9}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:31, 6 April 2021

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

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

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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