2020 AMC 12B Problems/Problem 8: Difference between revisions
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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> | This will clearly only yield real solutions when <math>x^{2020} \leq 1</math>, because it is always positive. | ||
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: | |||
<cmath>y^2-2y+1=(y-1)^2=0</cmath> | |||
This has only has solutions <math>1</math>, so <math>(\pm 1,1)</math> are solutions. | |||
Next, if <math>x=0</math>: | |||
<cmath>y^2-2y=0</cmath> | |||
Which has 2 solutions, so <math>(0,2)</math> and <math>(0,0)</math> | |||
These are the only 4 solutions, so <math>\boxed{D}</math> | |||
==See Also== | ==See Also== | ||
Revision as of 19:28, 7 February 2020
Problem
How many ordered pairs of integers 
satisfy the equation ![\[x^{2020}+y^2=2y\]](http://latex.artofproblemsolving.com/6/7/5/675a2782e7f97942c051c125abe88fb075b0d07c.png)
Solution
Set it up as a quadratic in terms of y:
![\[y^2-2y+x^{2020}=0\]](http://latex.artofproblemsolving.com/5/7/3/57302fcd989d1abb3190666693fe5a475f8fdd1c.png)
Then the discriminant is
![\[\Delta = 4-4x^{2020}\]](http://latex.artofproblemsolving.com/2/f/a/2fa32c16b6896590c31f648e13db1aafd9439433.png)
This will clearly only yield real solutions when 
, because it is always positive.
Then 
. Checking each one:
and 
are the same when raised to the 2020th power:
![\[y^2-2y+1=(y-1)^2=0\]](http://latex.artofproblemsolving.com/0/e/a/0eab1c8bfe244a00f60be70bfd800c5b1508a574.png)
This has only has solutions , so 
are solutions.
Next, if 
:
![\[y^2-2y=0\]](http://latex.artofproblemsolving.com/e/b/f/ebf1c43ea6bebf5bd1f2288237ac84584bbbbe85.png)
Which has 2 solutions, so 
and 
These are the only 4 solutions, so 
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America. 
