Difference between revisions of "2020 AMC 12B Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
− | 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{D}</math> | + | 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> |
==See Also== | ==See Also== |
Revision as of 23:00, 7 February 2020
Contents
Problem
How many ordered pairs of integers satisfy the equation
Solution
Set it up as a quadratic in terms of y:
Then the discriminant is
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:
This has only has solutions
, so
are solutions.
Next, if
:
Which has 2 solutions, so
and
These are the only 4 solutions, so
Solution 2
Move the term to the other side to get
. Because
for all
, then
. If
or
, the right side is
and therefore
. When
, the right side become
, therefore
. Our solutions are
,
,
,
. There are
solutions, so the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.