Difference between revisions of "2020 AMC 10B Problems/Problem 9"
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+ | ==Solution 3, x first== | ||
+ | Set it up as a quadratic in terms of y: | ||
+ | <cmath>y^2-2y+x^{2020}=0</cmath> | ||
+ | Then the discriminant is | ||
+ | <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. | ||
+ | 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{\textbf{(D) } 4}</math> | ||
+ | |||
+ | ==Solution 4, y first== | ||
+ | |||
+ | 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/ | + | https://youtu.be/6ujfjGLzVoE |
~IceMatrix | ~IceMatrix | ||
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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
Contents
Problem
How many ordered pairs of integers satisfy the equation
Solution
Rearranging the terms and and completing the square for yields the result . Then, notice that can only be , and because any value of that is greater than 1 will cause the term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a total of ordered pairs.
Solution 2
Bringing all of the terms to the LHS, we see a quadratic equation in terms of . Applying the quadratic formula, we get In order for to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, must be nonnegative. Therefore, Here, we see that we must split the inequality into a compound, resulting in .
The only integers that satisfy this are . Plugging these values back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .
Thus, the answer is .
~Tiblis
Solution 3, x first
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 4, y first
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
Video Solution
~IceMatrix
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.