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| − | ==Problem 12==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_1]] |
| − | Which of the following conditions is sufficient to guarantee that integers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the equation
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| − | <cmath>x(x-y)+y(y-z)+z(z-x) = 1?</cmath><math>\textbf{(A)}\: x>y</math> and <math>y=z</math>
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| − | <math>\textbf{(B)}\: x=y-1</math> and <math>y=z-1</math>
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| − | <math>\textbf{(C)} \: x=z+1</math> and <math>y=x+1</math>
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| − | <math>\textbf{(D)} \: x=z</math> and <math>y-1=x</math>
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| − | <math>\textbf{(E)} \: x+y+z=1</math>
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| − | ==Solution 1==
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| − | Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.
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| − | We have <math>y=x+1, z=x</math>, so
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| − | <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>
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| − | Our answer is <math>\textbf{(D)}</math>.
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| − | ~kingofpineapplz
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
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| − | {{MAA Notice}}
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