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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 7"

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<math>\textbf{(E)} \: x+y+z=1</math>
 
<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
  
 
==Solution 1 (Bash) ==
 
==Solution 1 (Bash) ==

Revision as of 22:31, 23 November 2021

Problem

Which of the following conditions is sufficient to guarantee that integers $x$, $y$, and $z$ satisfy the equation \[x(x-y)+y(y-z)+z(z-x) = 1?\]

$\textbf{(A)} \: x>y$ and $y=z$

$\textbf{(B)} \: x=y-1$ and $y=z-1$

$\textbf{(C)} \: x=z+1$ and $y=x+1$

$\textbf{(D)} \: x=z$ and $y-1=x$

$\textbf{(E)} \: x+y+z=1$

Solution 1

Plugging in every choice, we see that choice $\textbf{(D)}$ works.


We have $y=x+1, z=x$, so \[x(x-y)+y(y-z)+z(z-x)=x(x-(x+1))+(x+1)((x+1)-x)+x(x-x)=x(-1)+(x+1)(1)=1.\] Our answer is $\textbf{(D)}$.

~kingofpineapplz

Solution 1 (Bash)

Just plug in all these options one by one, and one sees that all but $D$ fails to satisfy the equation.

For $D$, substitute $z=x$ and $y=x+1$:

$LHS=x(x-(x+1))+(x+1)(x+1-x)+x(x-x)=(-x)+(x+1)=1=RHS$

Hence the answer is $\boxed{\textbf{(D)}}.$

~Wilhelm Z

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