2021 Fall AMC 12B Problems/Problem 7

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 $y-1=x$

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

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