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

Revision as of 21:57, 23 November 2021

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

@@ Line 1: / Line 1: @@
+==Problem==
+Which of the following conditions is sufficient to guarantee that integers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the equation
+<cmath>x(x-y)+y(y-z)+z(z-x) = 1?</cmath>
+<math>\textbf{(A)} \: x>y</math> and <math>y=z</math>
+<math>\textbf{(B)} \: x=y-1</math> and <math>y=z-1</math>
+<math>\textbf{(C)} \: x=z+1</math> and <math>y=x+1</math>
+<math>\textbf{(D)} \: x=z</math> and <math>y-1=x</math>
+<math>\textbf{(E)} \: x+y+z=1</math>
 ==Solution 1 (Bash) ==