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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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==Solution 1 (Bash) ==
−
Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.
 
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−
 
 
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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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==Solution 2 (Bash) ==
 
  
 
Just plug in all these options one by one, and one sees that all but <math>D</math> fails to satisfy the equation.
 
Just plug in all these options one by one, and one sees that all but <math>D</math> fails to satisfy the equation.
Line 35: Line 25:
  
 
~Wilhelm Z
 
~Wilhelm Z
 +
 +
==Solution 2==
 +
Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works.
 +
 +
 +
We have <math>y=x+1, z=x</math>, so
 +
<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>
 +
Our answer is <math>\textbf{(D)}</math>.
 +
 +
~kingofpineapplz

Revision as of 00:28, 24 November 2021

The following problem is from both the 2021 Fall AMC 10B #12 and 2021 Fall AMC 12B #7, so both problems redirect to this page.

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 (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

Solution 2

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

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