2021 Fall AMC 10B Problems/Problem 12

Problem 12

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

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2021 Fall AMC 10B Problems/Problem 12

Problem 12

Solution 1

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