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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 12"

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==Problem 12==
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Which of the following conditions is sufficient to guarantee that integers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the equation
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<cmath>x(x-y)+y(y-z)+z(z-x) = 1?</cmath><math>\textbf{(A)}\: x>y</math> and <math>y=z</math>
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<math>\textbf{(B)}\: x=y-1</math> and <math>y=z-1</math>
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<math>\textbf{(C)} \: x=z+1</math> and <math>y=x+1</math>
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<math>\textbf{(D)} \: x=z</math> and <math>y-1=x</math>
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<math>\textbf{(E)} \: x+y+z=1</math>
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[[2021 Fall AMC 10B Problems/Problem 12|Solution]]
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==Solution 1==
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Plugging in every choice, we see that choice <math>\textbf{(D)}</math> works. 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
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=13|num-b=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:15, 23 November 2021

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

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

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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