Difference between revisions of "2021 Fall AMC 10B Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works. We have <math>y=x+1, z=x</math>, so | Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works. We have <math>y=x+1, z=x</math>, so | ||
| − | <cmath>x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1</cmath> | + | <cmath>x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.</cmath> |
Our answer is <math>\textbf{(D)}</math>. | Our answer is <math>\textbf{(D)}</math>. | ||
~kingofpineapplz | ~kingofpineapplz | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 07:15, 23 November 2021
Problem 12
Which of the following conditions is sufficient to guarantee that integers 
, 
, and 
satisfy the equation
![\[x(x-y)+y(y-z)+z(z-x) = 1?\]](http://latex.artofproblemsolving.com/e/d/4/ed4334b3ae76f385d27c6bb80a0dbc73874bdf7d.png)
and 
and 
and 
and 
Solution 1
Plugging in every choice, we see that choice 
works. We have 
, so
![\[x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.\]](http://latex.artofproblemsolving.com/d/6/8/d6872ccc0bf390a8b408b95fa4c4fe66335d95e1.png)
Our answer is .
~kingofpineapplz
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
