Difference between revisions of "2020 AMC 10B Problems/Problem 9"
Advancedjus (talk | contribs) (→Solution) |
(→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | 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>(0,2)</math>, <math>(1,-1)</math>, and <math>(-1,-1)</math>, | + | 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>(0,2)</math>, <math>(1,-1)</math>, and <math>(-1,-1)</math>. Eliminating the extraneous roots <math>(1,-1)</math> and <math>(-1,-1)</math> gives a total of <math>\boxed{\textbf{(B) }2}</math> ordered pairs. |
==Video Solution== | ==Video Solution== |
Revision as of 18:58, 7 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
. Eliminating the extraneous roots
and
gives a total of
ordered pairs.
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.